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IN MEMORIAM
FLORIAN CAJORI
4^i^^a ^ (Sy^
WENTWORTH'S "^
PLANE GEOMETRY
REVISED BY
GEORGE WENTWORTH
AND
DAVID EUGENE SMITH
GINN AND COMPANY
BOSTON ■ NEW YORK • CHICAGO • LONDON
COPYRIGHT, 1888, 1899, BY G. A. WENT WORTH
COPYRIGHT, 1910, BY GEORGE WENTWORTH
AND DAVID EUGENE SMITH
ENTERED AT STATIONERS' HALL
ALL RIGHTS RESERVED
910.7
GINN AND COMPANY • PRO
PRIETORS • BOSTON • U.S.A.
b
^
I^Ai/^
PEEFACE
Long after the death of Robert Recorde, England's first
great writer of textbooks, the preface of a new edition of
one of his works contained the appreciative statement that
the book was " entail'd upon the People, ratified and sign'd by
the approbation of Time.'' The language of this sentiment
sounds quaint, but the noble tribute is as impressive today
as when first put in print two hundred fifty years ago.
With equal truth these words may be applied to the Geome>
try written by George A. Wentworth. For a generation it
has been the leading textbook on the subject in America. It
set a standard for usability that every subsequent writer upon
geometry has tried to follow, and the number of pupils who
have testified to its excellence has run well into the millions.
In undertaking to prepare a revision of this work, the authors
have been guided by certain welldefined principles, based upon
an extended investigation of the needs of the schools and upon
a study of all that is best in the recent literature of the sub
ject. The effects of these principles they feel should be sum
marized for the purpose of calling the attention of the wide
circle of friends of the Wentworth text to the points of simi
larity and of difference in the editions.
1. Every effort has. been made not only to preserve but to
improve upon the simplicity of treatment, the clearness of
expression, and the symmetry of page that have characterized
the successive editions of the Wentworth Geometry. It has
been the purpose to prepare a book that should do even more
than maintain the traditions this work has fostered.
rft\£y^ €\9%
iv PLANE GEOMETRY
2. The proofs have been given substantially in full, tOv.the
end that the pupil may always have before him a model for
his independent treatment of the exercises.
3. The sequence of propositions has been improved in sev
eral respects, notably in the treatment of parallels.
4. To meet a general demand, the number of propositions
has been decreased so as to include only the great basal theo
rems and problems. A little of the less important material
has been placed in the Appendix, to be used or not as circum
stances demand.
5. The exercises, in some respects the most important part
of a course in geometry, have been rendered more dignified in
appearance and have been improved in content. The number
of simple exercises has been greatly increased, while the diffi
cult puzzle is much less in evidence than in most American
textbooks. The exercises are systematically grouped, appear
ing in full pages, in large type, at frequent intervals. They
are not all intended for one class, but are so numerous as to
allow the teacher to make selections from year to year.
6. The introduction has been made as concrete as is reason
able. Definitions have been postponed until they are actually
needed, only wellrecognized terms have been employed, the
pupil is initiated at once into the practical use of the instru
ments, some of the reasons for studying geometry are early
shown in an interesting way, and correlation is made with
the simple algebra already studied.
The authors are indebted to many friends of the Wentworth
Geometry for assistance and encouragement in the labor of pre
paring this edition, and they will welcome any further sugges
tions for improvement from any of their readers.
GEORGE WENTWORTH
DAVID EUGENE SMITH
CONTENTS
Page
INTRODUCTION 1
BOOK I. RECTILINEAR FIGURES 25
Triangles 26
Parallel Lines 46
Quadrilaterals ......... 59
Polygons 68
Loci 73
BOOK II. THE CIRCLE 93
Theorems 94
Problems 126
BOOK III. PROPORTION. SIMILAR POLYGONS . . .151
Theorems 152
Problems 182
BOOK IV. AREAS OF POLYGONS 191
Theorems 192
Problems 214
BOOK V. REGULAR POLYGONS AND CIRCLES . . 227
Theorems 228
Problems 242
APPENDIX 261
Symmetry 261
Maxima and Minima ........ 265
Recreations 273
History of Geometry ........ 277
INDEX 283
V
SYMBOLS AND ABBREVIATIONS
= equals, equal, equal to,
Adj.
adjacent.
is equal to, or
Alt.
alternate.
is equivalent to.
Ax.
axiom.
> is greater than.
Const.
construction.
< is less than.
Cor.
corollary.
II parallel.
Def.
definition.
_L perpendicular.
Ex.
exercise.
Z angle.
Ext.
exterior.
A triangle.
Fig.
figure.
O parallelogram.
Hyp.
hypothesis.
□ rectangle.
Iden.
identity.
O circle.
Int.
interior.
st. straight.
Post.
postulate.
rt. right.
Prob.
problem.
•.• since.
Prop.
proposition.
.'. therefore.
Sup.
supplementary.
These symbols take the plural form when necessary, as in the case of
The symbols +, — , x, ^ are used as in algebra.
There is no generally accepted symbol for "is congruent to," and the
words are used in this book. Some teachers use = or =, and some use
= , but the sign of equality is more commonly employed, the context
telling whether equality, equivalence, or congruence is to be understood.
Q. E. D. is an abbreviation that has long been used in geometry for
the Latin words quod erat demonstrandum, "which was to be proved."
Q. E. F. stands for quod erat faciendum, "which was to be done."
PLANE GEOMETRY
INTRODUCTION
1. The Nature of Arithmetic. In arithmetic we study compu
tation, the working with numbers. We may have a formula
expressed in algebraic symbols, such as a = hh, where a may
stand for the area of a rectangle, and h and h respectively for
the number of units of length in the base and height ; but the
actual computation involved in applying such formula to a
particular case is part of arithmetic.
2. The Nature of Algebra. In algebra we generalize the
arithmetic, and instead of saying that the area of a rectangle
with base 4 in. and height 2 in. is 4 x 2 sq. in., we express a
general law by saying that a = hh. In arithmetic we may have
an equality, like 2 x 16 +17= 49, but in algebra we make much
use of equations, like 2 cc f 17 = 49. Algebra, therefore, is a
generalized arithmetic.
3. The Nature of Geometry. We are now about to begin another
branch of mathematics, one not chiefly relating to numbers
although it uses numbers, and not primarily devoted to equa
tions although using them, but one that is concerned principally
with the study of forms, such as triangles, parallelograms, and
circles. Many facts that are stated in arithmetic and algebra
are proved in geometry. For example, in geometry it is proved
that the square on the hypotenuse of a right triangle equals
the sum of the squares on the other two sides, and that the
circumference of a circle equals 3.1416 times the diameter,
1
PLANE GEOMETKY
4. Solid. The block here represented is called a solid; it
is a limited portion of space filled with matter. In geometry,
however, we have nothing to do with the matter of which a
body is composed; we study simply its sha^je and size, as in
the second figure.
That is, a physical solid can be touched and handled ; a geometric
solid is the space that a physical solid is conceived to occupy. For
example, a stick is a physical solid ; but if we put it into wet plaster, and
then remove it, the hole that is left may be thought of as a geometric
solid although it is filled with air.
5. Geometric Solid. A limited portion of space is called a
geometric solid.
6. Dimensions. The block represented in § 4 extends in three
principal directions :
(1) From left to right, that is, from A to D]
(2) From back to front, that is, from A to B]
(3) From top to bottom, that is, from A to E.
These extensions are called the dimensions of the block, and
are named in the order given, length, breadth (or width), and
thickness (height, altitude, or depth). Similarly, we may say
that every solid has three dimensions.
Very often a solid is of such shape that we cannot point out the length,
or distinguish it from the breadth or thickness, as an irregular block of
coal. In the case of a round ball, where the length, breadth, and thick
ness are all the same in extent, it is impossible to distinguish one dimen
sion from the others.
INTEODUCTIO>T 3
7. Surface. The block shown in § 4 has six flat faces, each
of which is called a surface. If the faces are made smooth by
polishing, so that when a straight edge is applied to any one
of them the straight edge in every part will touch the surface,
each face is called 2^ plane surface, or 2^ plane.
These surfaces are simply the boundaries of the solid. They have no
thickness, even as a colored light shining upon a piece of paper does not
make the paper thicker. A board may be planed thinner and thinner,
and then sandpapered still thinner, thus coming nearer and nearer to
representing what we think of as a geometric plane, but it is always a
solid bounded by surfaces.
That which has length and breadth without thickness is
called a surface.
8. Line. In the solid shown in § 4 we see that two adja
cent surfaces intersect in a line. A line is therefore simply
the boundary of a surface, and has neither breadth nor thickness.
That which has length without breadth or thickness is called
a line.
A telegraph wire, for example, is not a line. It is a solid. Even a
pencil mark has width and a very little thickness, so that it is also a solid.
But if we think of a wire as drawn out so that it becomes finer and finer,
it comes nearer and nearer to representing what we think of and speak
of as a geometric line.
9. Magnitudes. Solids, surfaces, and lines are called tnag
nitudes.
10. Point. In the solid shown in § 4 we see that when two
lines meet they meet in a point. A point is therefore simply
the boundary of a line, and has no length, no breadth, and
no thickness.
That which has only position, without length, breadth, or
thickness, is called di: point.
We may think of the extremity of a line as a point. We may also
think of the intersection of two lines as a point, and of the intersection
of two surfaces as a line.
4 PLANE GEOMETRY
11. Representing Points and Geometric Magnitudes. Although
we only imagine such geometric magnitudes as lines or planes,
we may represent them by pictures.
Thus we represent a point by a fine dot, and I \ 7
name it by a letter, as P in this figure. / /
We represent a hne by a fine mark, and name L '
it by letters placed at the ends, as ^5.
We represent a surface by its boundary lines, and name it by letters
placed at the corners or in some other convenient way, as A BCD.
We represent a solid by the boundary faces or by the lines bounding
the faces, as in § 4.
12. Generation of Geometric Magnitudes. We may think of
(1) A line as generated by a moving point ;
(2) A surface as generated by a moving line ;
(3) A solid as generated by a moving surface.
For example, as shown in the figure let the surface A BCD move to
the position WXYZ. Then
(1) A generates the line AW;
(2) AB generates the surface A WXB ; D
(3) ABCD generates the solid AY.
C ., Y
A
— >—
B
71
>■
Of course a point will not generate a line / ' /
by simply turning over, for this is not mo ^ ^ ^
tion for a point ; nor will a line generate a
surface by simply sliding along itself ; nor will a surface generate a solid
by simply sliding upon itself.
13. Geometric Figure. A point, a line, a surface, a solid, or
any combination of these, is called a geometric figure.
A geometric figure is generally called simply a figure.
14. Geometry. The science of geometric figures is called
geometry.
Plane geometry treats of figures that lie wholly in the same
plane, that is, of plane figures.
Solid geometry treats of figures that do not lie wholly in
the same plane.
INTEODUCTIOK
15. Straight Line. A line such that any part placed with its
ends on any other part must lie wholly in the line is called a
straight line.
For example, J.i> is a straight line, for if we take, say, a half inch of it,
and place it in any way on any other part of AB,
but so that its ends lie in AB^ then the whole of
the half inch of line will lie in AB. This is well shown by using tracing
paper. The word line used alone is understood to mean a straight line.
Part of a straight line is called a segment of the line. The term seg
ment is applied also to certain other magnitudes.
16. Equality of Lines. Two straightline segments that can
be placed one upon the other so that their extremities coin
cide are said to be equal.
In general, two geometric magnitudes are equal if they can be
made to coincide throughout their whole extent. We shall see later
that some figures that coincide are said to be congruent.
17. Broken Line. A line made up of
two or more different straight lines is
called a broken line.
For example, CD is a broken line.
18. Rectilinear Figure. A plane figure
bounded by a broken line is called a rec
tilinear figure.
For example, A BCD is a rectilinear figure.
19. Curve Line. A line no part of which
is straight is called a curiae line, or simply
a curve.
For example, EF is a curve line.
20. Curvilinear Figure. A plane figure formed
by a curve line is called a curvilinear figure.
For example, is a curvilinear figure with which
we are already familiar.
Some curvilinear figures are surfaces bounded by
curves and others are the curves themselves.
•O
6
PLANE GEOMETEY
21. Angle. The opening between two straight lines drawn
from the same point is called an angle.
Strictly speaking, this is a plane angle. We shall
find later that there are angles made by curve lines and
angles made by planes.
The two lines are called the sides of the angle, and q
the point of meeting is called the vertex.
An angle may be read by naming the letters desig
nating the sides, the vertex letter being between the
others, as the angle A OB. An angle may also be desig
nated by the vertex letter, as the angle O, or by a small
letter within, as the angle m. A curve is often drawn to show the par
ticular angle meant, as in angle m.
22. Size of Angle. The size of an angle depends upon the
amount of turning necessary to bring one side into the position
of the other.
One angle is greater
than another angle when
the amount of turning is
greater. Thus in these
compasses the first angle
is smaller than the second, which is also smaller than the third,
length of the sides has nothing to do with the size of the angle.
The
23. Equality of Angles. Two angles that can be placed one
upon the other so that their vertices coincide and the sides of
one lie along the sides of the other are said to ^b
be equal.
For example, the angles AOB and A'O'B' (read
"J. prime, prime, B prime") are equal. It is well
to illustrate this by tracing one on thin paper and
placing it upon the other. q'^c ^
24. Bisector. A point, a line, or a plane that divides a geo
metric magnitude into two equal parts is called a bisector of
the magnitude.
For example, M, the midpoint of the line AB^ A M B
is a bisector of the line.
INTRODUCTION
25. Adjacent Angles. Two angles that have the same vertex
and a common side between them are called adjacent angles.
For example, the angles AOB and BOC are
adjacent angles, and in §26 the angles AOB and
BOC are adjacent angles.
26. Right Angle. When one straight line
meets another straight line and makes the
adjacent angles equal, each angle is called a
right angle.
For examplie, angles A OB and BOCm this figure.
If CO is cut off, angle AOB is still a right angle. ^ ^ ^
27. Perpendicular. A straight line making a right angle with
another straight line is said to be ijeirpendicular to it.
Thus OB is perpendicular to CA^ and CA to OB. OB is also called a
perpendicular to CJ., and is called the foot of the perpendicular OB.
28. Triangle. A portion of a plane bounded by three straight
lines is called a triangle. C
The lines AB, BC, and CA are called the sides
of the triangle ABC^ and the sides taken together
form the perimeter. The points A, B, and C are
the vertices of the triangle, and the angles A, B, and C are the angles of
the triangle. The side AB upon which the triangle is supposed to rest
is the base of the triangle. Similarly for other plane figures.
29. Circle. A closed curve lying in a plane, and such that
all of its points are equally distant from a fixed point in the
plane, is called a circle.
The length of the circle is called the circumference.
The point from which all points on the circle are
equally distant is the center. Any portion of a circle
is an arc. A straight line from the center to the circle
is a radius. A straight line through the center, termi
nated at each end by the circle, is a diameter.
Formerly in elementary geometry circle was taken to mean the space
inclosed, and the bounding line was called the circumference. Modern
usage has conformed to the definition used in higher mathematics.
8
PLANE GEOMETRY
30. Instruments of Geometry. In geometry only two instru
ments are necessary besides pencil and paper. These are a
straight edge, or ruler, and a pair of compasses.
It is evident that all radii of the same circle are equal.
In the absence of compasses, and particularly for blackboard work, a
loop made of string may be used. For the accurate transfer of lengths,
however, compasses are desirable.
31. Exercises in using Instruments. The following simple
exercises are designed to accustom the pupil to the use of
instruments. No proofs are attempted, these coming later in
the course.
This section may be omitted if desired, without affecting the course.
*.CL
EXERCISE 1
1. From a given point on a given straight line required to
draw a perpendicular to the line.
Let AB he the given line and P be the
given point.
It is required to draw from P a line per
pendicular to AB.
With P as a center and any convenient
radius draw arcs cutting AB 2it X and Y.
With JT as a center and XY as a radius draw a circle, and with Y
as a center and the same radius draw another circle, and call one inter
section of the circles C.
With a straight edge draw a line from P to C, and this will be the
perpendicular required.
\X
Yl
INTRODUCTION
9
A X^'
2. From a given point outside a given straight line required
to let fall a perpendicular to the line. p
Let AB he the given straight line and P be the
given point.
It is required to draw from P a line perpen
dicular to AB.
With P as a center and any convenient radius
draw an arc cutting AB at X and Y.
With X as a center and any convenient radius
draw a circle, and with F as a center and the same
radius draw another circle, and call one intersection of the circles C.
With a straight edge draw a straight line from P to C, and this will
be the perpendicular required.
It is interesting to test the results in Exs. 1 and 2, by cutting the paper
and fitting the angles together.
S^'
I
3. Required to draw a triangle having two sides each equal
to a given line.
Let I be the given line.
It is required to draw a triangle having two sides
each equal to I.
With any center, as C, and a radius equal to I
draw an arc.
Join any two points on the arc, as A and J5,
with each other and with C by straight lines.
Then ABC is the triangle required.
^'~
4. Required to draw a triangle having its three sides each
equal to a given line. ^ ^ ^
Let AB he the given line.
It is required to draw a triangle having its three
sides each equal to AB.
With A as a center and ^Z? as a radius draw a
circle, and with B as a center and the same radius
draw another circle.
Join either intersection of the circles with A and B by straight lines.
Then ABC is the triangle required.
In such cases draw the arcs only long enough to show the point of
intersection.
10
PLANE GEOMETPvY
^
^s.O.'
5. Eequired to draw a triangle having its sides equal respec
tively to three given lines.
Let the three lines be I, m, and n.
What is now required ?
Upon any line mark off with the com
passes a linesegment AB equal to I.
With ^ as a center and m as a radius
draw a circle ; with 2^ as a center and
n as a radius draw a circle.
Drawee and 50. ^
Then ABC is the required triangle. n
6. From a given point on a given line required to draw a
line making an angle equal to a given angle.
X
^^
o
\c
IM
Let P be the given point on the given line PQ, and let angle AOB be
the given angle.
What is now required ?
With as a center and any radius draw an arc cutting ^ O at C and
BO at D.
With P as a center and OC as a radius draw an arc cutting PQ at M.
With Jlf as a center and the straight line joining C and D as a radius
draw an arc cutting the arc just drawn at N, and draw PN.
Then angle MPN is the required angle.
7. Eequired to bisect a given straight line.
Let AB be the given line.
It is required to bisect it.
With J. as a center and ^B as a radius draw a circle,
and with B as a center and the same radius draw a circle.
Call the two intersections of the circles X and Y.
Draw the straight line XY.
Then XY bisects the line AB a^t the point of inter
section M.
^C
M
r
INTKODUCTION
11
8. Required to bisect a given angle.
Let A OB be the given angle.
It is required to bisect it.
With O as a center and any convenient radius
draw an arc cutting OA at X and OB at Y.
With X as a center and a line joining X and
F as a radius draw a circle, and with F as a
center and the same radius draw a circle, and call one point of inter
section of the circles P.
Draw the straight line OP.
Then OP is the required bisector.
9. By the use of compasses and ruler draw the following
figures : ^^ ^
The dotted lines show how to fix the points needed in drawing the
figure, and they may be erased after the figure is completed. In general,
in geometry, auxiliary lines (those needed only as aids) are indicated by
dotted lines.
10. By the use of compasses and ruler draw the following
figures :
It is apparent from the figures in Exs. 9 and 10 that the radius of
the circle may be used in describing arcs that shall divide the circle into
six equal parts.
12
PLANE GEOMETRY
11. By the use of compasses and ruler draw the following
figures :
12. By the use of compasses and ruler draw the following
figures :
13. By the use of compasses and ruler draw the following
figures :
In such figures artistic patterns may be made by coloring various
portions of the drawings. In this way designs are made for stainedglass
windows, for oilcloth, for colored tiles, and for other decorations.
14. Draw a triangle of which each side is li in.
15. Draw two lines bisecting each other at right angles.
INTRODUCTION 13
16. Bisect each of the four right angles formed by two lines
bisecting each other at right angles.
17. Draw a line 1^ in. long and divide it into eighths of an
inch, using the ruler. Then with the compasses draw this
figure.
It is easily shown, when we come to
the measurement of the circle, that these
two curve lines divide the space inclosed
by the circle into parts that are exactly
equal to one another.
By continuing each semicircle to
make a complete circle another inter
esting figure is formed. Other similar
designs are easily invented, and students
should be encouraged to make such
original designs.
18. In planning a Gothic window this drawing is needed.
The arc BC is drawn with ^ as a center q
and vlJ5 as a radius. The small arches
are described with A, Z), and B as centers
and ^i) as a radius. The center P is found
by taking A and B as centers and AE sls
a ladius. How may the points D, E, and
F be found ? Draw the figure. j_ p ^ E B
19. Draw a triangle of which each side is 1 in. Bisect each
side, and with the points of bisection as centers and with radii
^ in. long draw three circles.
20. A baseball diamond is a square 90 ft. on a side. Draw
the plan, using a scale of ^^ in. to a foot. Locate the pitcher
60 ft. from the home plate.
21. A man travels from A directly east 1 mi. to B. He then
turns and travels directly north 1 mi. to C. Draw the plan
and find by measurement the distance ^ C to the nearest quarter
of a mile. Use a scale of ^ in. to a mile.
14 PLANE GEOMETRY
22. A double tennis court is 78 ft. long and 36 ft. wide. The
net is placed 39 ft. from each end and the service lines 18 ft.
from each end. Draw the plan, using a scale of y^^ in. to a foot,
making the right angles as shown in Ex. 1. The accuracy of
the construction may be tested by measuring the diagonals,
which should be equal.
23. At the entrance to New York harbor is a gun having
a range of 12 mi. Draw a line inclosing the range of fire,
using a scale of ^^ in. to a mile.
24. Two forts are placed on opposite sides of a harbor
entrance, 13 mi. apart. Each has a gun having a range of
10 mi. Draw a plan showing the area exposed to the fire of
both guns, using a scale of ^\ in. to a mile.
25. Two forts, A and B, are placed on opposite sides of a
harbor entrance, 16 mi. apart. On an island in the harbor, 12
mi. from A and 11 mi. from B, is a fort C. The fort A has a
gun with a range of 12 mi., fort B one with a range of 11 mi.,
and fort C one with a range of 10 mi. Draw a plan of the
entrance to the harbor, showing the area exposed to the fire
of each gun.
26. A horse, tied by a rope 25 ft. long at the corner of a lot
50 ft. square, grazes over as much of the lot as possible. The
next day he is tied at the next corner, the third day at the
third corner, and the fourth day at the fourth corner. Draw
a plan showing the area over which he has grazed during the
four days, using a scale of ^ in. to 5 ft.
27. A gardener laid out a flower bed on the following plan :
He made a triangle ABC, 16 ft. on a side, and then bisected
two of the angles. From the point of intersection of the bi
sectors, P, he drew perpendiculars to the three sides of the
triangle, PX, PY, and PZ. Then he drew a circle with P as a
center and PX as a radius, and found that it just fitted in the
triangle. Draw the plan, using a scale of i in. to a foot.
IKTEODUCTIOK
15
32. Necessity for Proof. Although pUrt of geometry consists
in drawing figures, this is not the most important part. It is
essential to prove that the figures are what we claim them to be.
The danger of trusting to appearances is seen in Exercise 2.
EXERCISE 2
1. Estimate which is the longer line, AB or XF, and how
much longer. Then test your estimate by ^> <^B
measuring with the compasses or with a
piece of paper carefully marked.
2. Estimate which is the longer line, AB ot
CDj and how much longer. Then test your
estimate by measuring as in Ex. 1.
3 Look at this figure and state whether
AB and CD are both straight lines. If one
is not straight, which one is
it? Test your answer by us
ing a ruler or the folded edge
of a piece of paper.
4. Look at this figure and
state whether AB and CD are
the same distance apart at .4 and C as
at B and D. Then test your answer as
in Ex. 1.
5 . Look at this figure and state whether
AB will, if prolonged, lie on CD. Also
state whether WX will, if prolonged, lie
on YZ. Then test your answer
<
^ mmm^} ^
by laying a ruler along the lines. ^^
6. Look at this figure and state which
of the three lower lines is ^B prolonged.
Then test your answer by laying a ruler
along AB,
i
16 PLANE GEOMETRY
33. Straight Angle. When the sides of an angle extend in
opposite directions, so as to be in the same straight line, the
angle is called a straight angle. ^^
For example, the angle A OB, as shown in this ^
figure, is a straight angle. The angle BOA, below the line, is also a
straight angle.
34. Right Angle and Straight Angle. It follows from the
definition of right angle (§ 26) that a right angle is half of a
straight angle.
In like manner, it follows that a straight angle equals twice
a right angle.
35. Acute Angle. An angle less than a right angle is called
an aciite angle.
For example, the angle m, as shown in this figure, is
an acute angle.
36. Obtuse Angle. An angle greater than a right angle and
less than a straight angle is called an obtuse „
angle.
For example, the angle AOB, as shown in this . ,
figure, is an obtuse angle. ''^9^y
37. Reflex Angle. An angle greater than a straight angle
and less than two straight angles is called a reflex angle.
For example, the angle BOA, marked with a dotted curve line in the
figure in § 36, is a reflex angle.
When we speak of an angle formed by two given lines drawn from a
point we mean the smaller angle unless the contrary is stated.
38. Oblique Angles. Acute angles and obtuse angles are called
oblique angles.
The sides of oblique angles are said to be oblique to each
other, and are called oblique lines.
Evidently if we bisect a straight angle, we form two right angles ; if
we bisect a right angle or an obtuse angle, we form two acute angles ;
if we bisect a reflex angle, we form two obtuse angles.
INTEODUCTION
17
39. Generation of Angles. Suppose the line r to revolve from
the position OA about the point O as a vertex to the posi
tion OB. Then r describes or generates
the acute angle A OB, and, as we have seen
(§ 22) the size of the angle depends upon the
amount of rotation, the angle being greater
as the amount of turning is greater.
If r rotates still further, to the position OC, it
has then generated the right angle AOC and is
perpendicular to OA.
If r rotates still further, to the position OD, it has then generated the
obtuse angle A OB.
If r rotates to the position OE^ it has then generated the straight
angle AOE.
If r rotates to the position OF, it has then generated the reflex
angle A OF.
If r rotates still further, past 06? to the position OA again, it has
made a complete revolution and has generated two straight angles or
four right angles.
40. Sums and Differences of Magnitudes. If the straight line
AP has been generated by a point P , i___+__>P
moving from A to P, the segments ^
t
B CD
AB, BC, CD, and so on, having been generated in succession,
then we call AC the stem oi AB and BC. That is,
AC = AB\ BC, whence AC — BC = AB.
If the angle A OD has been generated by the  (^
line OA revolving about as a vertex from
the position OA, the angles AOB, BOC, and
COD having been generated in succession,
then we call angle AOC the sum of angles
AOB and BOC. That is, considering angles,
.4 OC = ^ 0^ + BOC, whence AOC  BOC = A OB.
In the same way that we may have the sum or the difference of lines
or of angles we may have the sum or the difference of surfaces or of solids.
18 PLANE GEOMETRY
41. Perigon. The whole angular space in a plane about a
point is called a perigon.
It therefore follows that a perigon equals the sum of two straight
angles or the sum of four right angles.
42. Complements, Supplements, and Conjugates, If the sum
of two angles is a right angle, each angle is called the comple
ment of the other.
If the sum of two angles is a straight angle, each angle is
called the supplement of the other.
If the sum of two angles is a perigon, each
angle is called the conjugate of the other.
Thus, with respect to angle AOB,
the complement is angle BOC,
the supplement is angle J50D,
the conjugate is angle BOA (reflex).
43. Properties of Supplementary Angles. It is sufficiently evi
dent to be taken without proof that
1. The two adjacent angles which one straight line Tuakes with
another are together equal to a straight angle.
2. If the sum of two adjacent angles is a straight angle, their
exterior sides are in the sam,e straight line.
44. Angle Measure. Angles are measured by taking as a unit
•j^Q of a perigon. This unit is called a degree.
The degree is divided into 60 equal parts, called minutes, and
the minute into 60 equal parts, called seconds.
We write 5° 13' 12" for 5 degrees 13 minutes 12 seconds.
It is evident that a right angle equals 90°, a straight angle equals 180°,
and a perigon equals 360°.
45. Vertical Angles. When two angles have the same vertex,
and the sides of the one are prolongations of
the sides of the other, those angles are called \
vertical angles. ^\"
In the figure the angles x and z are vertical angles, \
as are also the angles w and y. \
INTRODUCTION 19
EXERCISE 3
1. Find the complement of 72°; of 65^30'; of 22° 20' 15".
2. What is the supplement of 45° ? of 120° ? of 145° 5' ?
of 22° 20' 15" ?
3. What is the conjugate of 240° ? of 280° ? of 312° 10' 40" ?
4. The complement of a certain angle x is 2x.
How many degrees are there in x ?
yx
5. The complement of a certain angle ic is 3ic. How many
degrees are there in a? ?
6. What is the angle of which the complement is four times
the angle itself ?
7. The supplement of a certain angle a; is 5 ic. ^^
How many degrees are there in ic ? ^xy^
8. The supplement of a certain angle x is 14 x. How many
degrees are there in cc ?
9. What is the angle of which the supplement equals half
of the angle itself ?
10. How many degrees in an ahgle that equals its own com
plement ? in one that equals its own supplement ?
11. The conjugate of a certain angle ic is fee. A ^ A
How many degrees are there in cc ? y^
12. The conjugate of a certain angle x is \x. How many
degrees are there in cc ?
13. How many degrees in an angle that equals a third of its
own conjugate? in one that equals its own conjugate?
14. Find two angles, x and y, such that their sum is 90° and
their (difference is 10°.
15. Find two complementary angles such that their differ
ence is 30°.
16. Find two supplementary angles such that one is 20°
greater than the other.
20 PLANE GEOMETRY
17. The angles x and y are conjugate angles, and their differ
ence is a straight angle. How many degrees are there in each ?
18. The angles x and y are conjugate angles, and their differ
ence is zero. How many degrees are there in each ?
19. Of two complementary angles one is four fifths of the
other. How many degrees are there in each ?
20. Of two supplementary angles one is five times the other.
How many degrees are there in each ?
21. How many degrees are there in the smaller angle formed
by the hands of a clock at 5 o'clock ?
22. How many degrees are there in the smaller angle formed
by the hands of a clock at 10 o'clock ? b
23. In this figure, if angle A OB is 38°, how y^
many degrees in angle BOC ? How many in Xo
angle COD? How many in angle DO A ? j{
24. In the same figure, if angle AOB is equal to a third of
angle BOC, how many degrees in each of the four angles ?
25. In the angles of this figure, \iw = 2x, how
many degrees in each ? How many degrees iny? \
How many degrees in ^ ? A^
26. Find the angle whose complement de \
creased by 30° equals the angle itself.
27. Find the angle whose complement divided by 2 equals
the angle itself.
28. Draw a figure to show that if two adjacent angles have
their exterior sides in the same straight line, their sum is a
straight angle.
29. Draw a figure to show that the sum of all the angles
on the same side of a straight line, at a given point, is equal
to two right angles.
30. Draw a figure to show that the complements of equal
angles are equal.
INTRODUCTION 21
46. Axiom. A general statement admitted without proof to
be true is called an axiom.
For example, it is stated in algebra that "if equals are added to
equals the sums are equal." This is so simple that it is generally accepted
without proof. It is therefore an axiom.
47. Postulate. In geometry a geometric statement admitted
without proof to be true is called a postulate.
For example, it is so evident that all straight angles are equal, that
this statement is a postulate. It is also evident that a straight line may
be drawn and that a circle may be described, and these statements are
therefore postulates of geometry.
Axioms are therefore general mathematical assumptions, while geo
metric postulates are the assumptions peculiar to geometry. Postulates
and axioms are the assumptions upon which the whole science of mathe
matics rests.
48. Theorem. A statement to be proved is called a theorem.
For example, it is stated in arithmetic that the square on the hypote
nuse of a right triangle equals the sum of the squares on the other two
sides. This statement is a theorem to be proved in geometiy.
49. Problem. A construction to be made so that it shall
satisfy certain given conditions is called a prohlem.
For example, required to construct a triangle all of whose sides shall
be equal. This construction was made in § 31, Ex. 4, and later it will be
proved that the construction was correct.
50. Proposition. A statement of a theorem to be proved or
a problem to be solved is called a proposition.
In geometry, therefore, a proposition is either a theorem or a problem.
We shall find that most of the propositions at first are theorems. After
we have proved a number of theorems so that we can prove that the
solutions of problems are correct, we shall solve some problems.
51. Corollary. A truth that follows from another with little
or no proof is called 2, corollary.
For example, since we admit that all straight angles are equal, it follows
as a corollary that all right angles are equal, since a right angle is half
of a straight angle.
22 PLANE GEOMETRY
52. Axioms. The following are the most important axioms
used in geometry:
1. If equals are added to equals the sums are equal.
2. If equals are subtracted from equals the remainders are equal.
3. If equals are m^ultvplied by equals the products are equal.
4. If equals are divided by equals the quotients are equal.
In division the divisor is never zero,
5. Like powers or like p)ositive roots of equals are equal.
We learn from algebra that the square root of 4 is + 2 or — 2, but of
course these are not equal. In geometry we shall use only the positive roots.
6. If unequals are operated on by positive equals in the same
way, the results are unequal in the same order.
Taking a>h and taking x and y as equal positive quantities, this
axiom states that
a4x>& + 2/, a — x>b — y, ax>by, >, etc.
X y
7. If unequals are added to unequals in the same order, the
sum,s are unequal in the same order ; if unequals are subtracted
from equals the remainders are unequal in the reverse order.
If a > 6, c > d, and x = y, then a + c>h ^ d, and x — a <y — h.
8. Quantities that are equal to the same quantity or to equal
quantities are equal to each other.
9. A quantity may be substituted for its equal in an equation
or in an inequality.
Thus if X = & and \i a { x = c, then a + 6 = c ; and if a + x > c, then
a + 6 > c. Axiom 8 is used so often that it is stated separately, although
it is really included in Axiom 9.
10. If the first of three quantities is greater than the second,
and the second is greater than the third, then the first is greater
than the third.
Thus if a > 6, and if 6 > c, then a> c.
11. The whole is greater than any of its parts, and is equal
to the sum of all of its pjarts.
INTEODUCTION 23
53. Postulates. The following are among the most impor
tant postulates used in geometry. Others will be . introduced
as needed.
1. One straight line and only one can hfi drawn through two
given points.
2. A straight line may be produced to any required length.
To produce A B means to extend it through B ;
to produce BA means to extend it through A.
3. A straight line is the shortest path between two p)oints.
4. A circle may be described with any given j^oint as a center
and any given line as a radices.
5. Any figure may be 7noved from one 2jl(^(^(i to another with
out altering its size or shajje.
6. All straight angles are equal.
54. Corollary 1. Ttvo p>oints determine a straight line.
This is only a brief way of stating Postulate 1.
55. Corollary 2. Two straight lines can intersect in only
one point.
For if they had two points in common they would coincide (Post. 1).
56. Corollary 3. All right angles are equal.
For all straight angles are equal (Post. 6), and a straight angle (§ 34)
is twice a right angle. Hence Axiom 4 applies.
57. Corollary 4. From a given point in a given line only
one perpendicular can be drawn to the line. C b
For if there could be two perpendiculars
to DA at 0, as OB and OC, we should have
angles AOB and AOC both right angles, which
is impossible (§ 56). j)
58. Corollary 5. Equal angles have equal complements,
equal supplements, and equal conjugates.
59. Corollary 6. The greater of two angles has the less
complement, the less suj^plcTnent, and the less conjugate.
24 PLANE GEOMETRY
EXERCISE 4
1. If 10° + Z ic = 27° 30', find the value of Z x.
2. If Z a; + 37° = I Z ic + 40°, find the value of Z x.
3. If Za: + Z^> = 5Z^, find the value of Zic.
4. lfZ.x\/a = ^/.a — Z.X, find the value of Z ic.
^mc? the value of /.x in each of the following equations :
5. Za;+13° = 39°. 10. Zic = 0.7Z;i; + 33°.
6. Zee 17° = 46°. 11. Za^ = 0.1Zic+18°. ^
7. 2 Zee ZicH 23°. 12. fZec = ^Zec + 2^°.
8. 5Zx = 2Zx421°. 13.  Zee = 0.1 Zee +14°.
9. 4Zee = iZee+70°. 14. f Za^ = JZee + 2°.
15. 12Za^+17°=9Z.r + 32°.
16. 5Za;22°30' = 2Zec+ll°.
17. 51°20'§Z:r = 5°l' + 3Za;.
18. 73° 21' 4"  Z X = 3° 3' 12" + 4 Z ee.
19. If ec + 20° = y and ?/ — 5° = 2 ee, what is the value of ec
and of ?/ ?
Find the value of x and of y in each of the following sets
of equations :
20. ee + ?/ = 45°, 23. ^ + 2^ = 21°,
xtj = S5°. ^ + 32/ = 26°15'.
21. xStj = 0'', 24. ee + 2/ = 9° 20' 15",
a; + 8?/=:80°. 2ee2/ = 12°25' 15".
22. 2ee + 2/ = 64°, 25. ee  2/ = 5'5",
See  2/ = 88°. 3ee + 4 1/ = 14° 50' 50".
26. If ec < 10° and y = 7° 30', what can be said as to the
value oi xiy?
27. In Ex. 26, what can be said as to the value oi x — y?
BOOK I
RECTILINEAR FIGURES
Proposition I. Theorem
60. If tioo lines intersect, the vertical angles are equal.
D^ A
Given the lines AC and BD intersecting at O.
To prove that AAOB = Z.COD.
Proof. Z .1 OB\A BOC = a st. Z. § 43
( The two adjacent angles which one straight line makes with another
are together equal to a straight angle.)
Likewise Z50C Z COT) = a st. Z. §43
..ZAOB + Z BOC = Z BOC \ Z COD. Post. 6
(All straight angles are equal.)
.\ZAOB = ZCOD. Ax. 2
(If equals are subtracted from equals the remainders are equal.) Q.E.D.
61. Nature of a Proof. From Prop. I it is seen that a tlieorem
has (1) certain things given; (2) a definite thing to he p)roved ;
(3) a proof, consisting of definite statements, each supported
by the authority of a definition, an axiom, a postulate, or some
proposition previously proved.
25
26 BOOK I. PLANE GEOMETEY
62. Triangles classified as to Sides. A triangle is said to be
scalene when no two of its sides are equal ;
isosceles when two of its sides are equal ;
equilateral when all of its sides are equal.
Scalene Isosceles Equilateral
63. Triangles classified as to Angles. A triangle is said to be
right when one of its angles is a right angle ;
obtuse when one of its angles is an obtuse angle ;
acute when all of its angles are acute angles ;
equiangular when all of its angles are equal.
Right Obtuse Acute Equiangular
64. Corresponding Angles and Sides. If two triangles have
the angles of the one respectively equal to the angles of the
other, the equal angles are called corresponding angles, and the
sides opposite these angles are called corresponding sides.
Corresponding parts are also called homologous parts.
65. Square. A rectilinear figure having four equal sides and
four right angles is called a square.
66. Congruent. If two figures can be made to coincide in all
their parts, they are said to be congruent.
67. Corollary. Corresponding parts of congruent figures
are equal.
When equal figures are necessarily congruent, as in the case of angles
or straight lines, the word equal is used. For symbols see page vi.
TKIANGLES 27
Proposition II. Theorem
68. Tivo triangles are congruent if tivo sides and the
included angle of the one are equal respectively to tivo
sides and the included angle of the other.
Given the triangles ABC and XYZ, with AB equal to XYy AC
equal to XZ, and the angle A equal to the angle X
To prove that A ABC is congruent to AXYZ.
Proof. Place the A ABC upon the AXYZ so that A shall
fall on X and AB shall fall along AT. Post. 5
{Any figure may be moved from one place to another without
altering its size or shape.)
Then B will fall on F,
{For AB is given equal to XY.)
^C will fall along XZ,
{For ZA is given equal to ZX.)
and C will fall on Z.
{For AC is given equal to XZ.)
.'. CB will coincide with ZY. Post. 1
{One straight line and only one can he drawn through two given points.)
.'. the two A coincide and are congruent, by § 66. q.e.d.
69. Corollary. Two right triangles are congruent if the
sides of the right angles are equal respectively.
The right angles are equal {§ 56). How does Prop. II apply ?
28
BOOK I. PLANE GEOMETEY
EXERCISE 5
1. In this figure if Z a = 53°, how many degrees are there
inZ^/? inZcc? inZ^?
2. In Ex. 1, if Z a were increased to 89°, what ^^^
would then be the size of A x, y, and z ?
3. In the square ABCD, prove that ^C = J5Z).
In ^ABC and BAB what two sides of the one are
known to be equal to what two sides of the other ?
How about the included angles ? "Write a complete
proof as in Prop. II.
4. If ABCD is a square and P is the mid
point of AB, prove that PC = PD.
What triangles should be proved congruent ? Can
this be done by Prop. II ? Write the proof.
5. How many degrees in an angle that equals a . p b
one fourth of its complement ? one tenth of its complement ?
6. How many degrees in an angle that equals twice its
supplement ? one third of its supplement ?
7. In the square ABCD the points P, Q,
i?, 5 bisect the consecutive sides. Prove that
PQ=::QR = RS=SP.
8. In the square ABCD the point P bisects
CD, and BM is made equal to AN, as shown in
this figure. Prove that PM = PN.
What two sides and included angle of one triangle
must be proved equal to what two sides and included N
angle of another triangle ?
9. Prove that to determine the distance AB
across a pond one may sight from A across a
post P, place a stake at A' making PA' = AP,
then sight along BP making PB' = BP, and
finally measure A'B'. ^^
TRIANGLES
29
70. Drawing the Figures. Directions have already been given
(§ 31) for drawing the most common geometric figures. For
example, in Prop. II the complete work of drawing AXYZ
so that XY=AB, Z.X = /.A, and XZ = AC, is indicated in
the following figures, the construction lines being dotted^ as is
always the case in this book.
It is desirable to construct such figures accurately, employing com
passes and ruler until such time as the use of these instruments is
thoroughly understood. Eventually, however, the figures should be
rapidly but neatly draw^n, freehand or with the aid of the ruler, as
the mathematician usually makes his figures.
71. Designating Corresponding Sides and Angles. It is helpful
in propositions concerning equality of figures to check the equal
parts so that the eye can follow the proof more easily. Thus it
would be convenient to represent the above figures as follows :
C Z
A • B X Y
Here AB and XFhave one check, AC and XZ two checks, and the
equal angles A and X are marked by curved arrows.
If a figure is very complicated, there is sometimes an advan
tage in using colored crayons or colored pencils, but otherwise
this expedient is of little value.
While such figures have some attraction for the eye they are not gen
erally used in practice, one reason being that the student rarely has a
supply of colored pencils at hand when studying by himself.
80 BOOK I. PLANE GEOMETRY
Proposition III. Theorem
72. Two triangles are congruent if two angles and
the included side of the one are equal respectively to
tivo angles and the included side of the other.
T
Given the triangles ABC and XYZ, with angle A equal to angle X,
angle B equal to angle F, and with AB equal to XY.
To prove that A ABC is congruent to AXYZ.
Proof. Place the A ABC upon the AXYZ so that AB shall
coincide with its equal, XY. Post. 5
{Any figure may be moved from one place to another without
altering its size or shape.)
Then AC will fall along XZ and BC along YZ.
{For it is given that ZA = ZX and ZB = ZY.)
.'. C will fall on Z. § 55
{Two straight lines can intersect in only one point.)
.'. the two A are congruent. § 66
{If two figures can he made to coincide in all their parts, they
are said to be congruent.) Q.E.D.
73. Hypothesis. A supposition made in an argument is called
an h7/j)othesls.
Thus, where it is said that ZA = ZX and ZB = ZY, yve might say
that this is true "by hypothesis," instead of saying that Z^ is given
equal to Z X, and Z jB is given equal to Z Y. The word is generally
used, however, for an assumption made somewhere in the proof.
TRIANGLES
31
M
EXERCISE 6
1. In the square ABCD the point 7^ bisects CD, and PQ and
PR are drawn so that Z QPC = 30° and Z RPQ = 120°. Prove
that PQ = PR.
If ZQPC = 30° and ZRPQ = 120°, what does ZDPR
equal ?
In the two triangles what parts are respectively
equal, and why ?
Write the proof in full. A
2. In this figure prove that if CAT bisects Z .1 CB and is also
perpendicular to AB, the triangle ABC is isosceles. ^
In i^AMC and BMC are two angles of the one respec
tively equal to two angles of the other ? Why ?
The two triangles have one common side.
Write the proof in full.
3. In the triangle ABC, AC = BC and CM bisects the angle C.
Prove that CM bisects the base AB.
4. The triangle ABC has Z.A equal to Z.B.
The point P bisects AB, and the lines PM
and PN are drawn so that Z.BPM=Z.NPA.
Prove t\\2it BM = AN.
5. The triangle ABC has Z .1 =Z.B. The lines
AP and BQ are so drawn that ZBAP = Z.QBA.
Prove that AP = BQ.
6. Wishing to measure the distance across a river, some
boys sighted from A to a point P. p
They then turned and measured AB
at right angles to AP. They placed
a stake at 0, halfway from A to B,
and drew a perpendicular to AB at B.
They placed a stake at C, on this
perpendicular, and in line with and P. They then found
the width by measuring BC. Prove that they were right
32 BOOK I. PLANE GEOMETRY
Proposition IV. Theorem
74. In an isosceles triangle the angles opposite the
equal sides are equal.
c
Given the isosceles triangle ABC^ with AC equal to BC.
To prove that Z.A = /.B.
Proof. Suppose CD drawn so as to bisect Z A CB.
Then in the A ADC and BDC,
AC = BC, Given
CD = CD, Iden.
{That is, CD is common to the two triangles.)
and ZACD = ZDCB. Hyp.
{For CD bisects Z AC B.)
..A ADC is congruent to A BDC. § 68
( Two A are congruent if two sides and the included Z of the one are equal
respectively to two sides and the included Z of the other.)
,\ZA=ZB. §67
{Corresponding parts of congruent figures are equal.) Q. E. D.
This proposition has long been known as the Pons asinorum, or Bridge
of Fools (asses). It is attributed to Thales, a Greek philosopher.
In an isosceles triangle the side which is not one of the two equal
sides is called the base.
75. Corollary. An equilateral triangle is equiangular.
Is ah equilateral triangle a special kind of isosceles triangle ?
tria:kgles 33
EXERCISE 7
1. With the figure of Prop. IV, if AC = BC and CD bisects
Z C, prove that CD is ± to AB.
What angles must be proved to be right angles ? What
is a right angle ? Do these angles fulfill the require
ments of the definition ?
2. In the adjacent figure ^C = BC. Prove that ^
/in=^ Z. n.
3. In the following figure AC = BC and AD = BD. Prove that
Z.CBD=^ADAC.
What angles are equal by Prop. IV ? Then what
axiom applies ?
4. In the figure of Ex. 3 prove that if a line ^
is drawn from C to D, the A DBC is congruent
to the A DA C. b
5. Two isosceles triangles, ABC and ABD, are constructed
on the same side of the common base AB. Prove
thsA. Z.CBD = Z.DAC.
c
6. In the figure of Ex. 5 prove that a line
drawn through C and D bisects Z ADB.
What two triangles must be proved congruent ?
7. Inthisfigure.4C = 5Cand^P=BQ. Prove that PC =QC.
Also prove that Z MPC = Z CQM. C
8. In this figure, if AC = BC, AP = BQ,
and PM=QM, prove that CM is ± to PQ.
What angles must be proved to be right angles ? J^ ^ Q~^B
9. In this figure P, Q, and R are midpoints of the sides of
the equilateral triangle ABC. Prove that PQR is ^
an equilateral triangle. ^^ x g
Prove that /kAPR, BQP, and CRQ are congruent by
using two propositions already proved. a
34 BOOK I. PLANE GEOMETEY
Proposition V. Theorem
76. If hvo angles of a triangle are equal, the sides
opposite the equal angles are equal, and the triangle
is isosceles.
c c'
A' B'
Given the triangle ABC, with the angle A equal to the angle B.
To prove that AC = BC.
Proof. Suppose the second triangle A'B'C' to be an exact
reproduction of the given triangle ABC.
Turn the triangle A'B'C' over and place it upon ABC so
that B' shall fall on A and A' shall fall on B. Post. 5
Then B'A' will coincide with AB. Post. 1
Since ZA' = ZB', Given
and ZA=ZA', Hyp.
.'.ZA=ZB'. Ax. 8
.'.^'C will lie along .1(7.
Similarly A'C will lie along BC.
Therefore C will fall on both AC and BC, and hence at
their intersection. . .
.'.B'C' = AC.
But B'C' was made equal to BC,
.'. AC = BC, by Ax. 8. q.e.d.
77. Corollary. An equiangular triangle is equilateral.
TRIANGLES 35
78. Kinds of Proof. In the five propositions thus far proved
in the text two different kinds of proof have been seen :
(1) Synthetic. In Prop. I we put together some known truths
in order to obtain a new truth. Such a method of proof is known
as the synthetic method, and is the most common of all that are
used in geometry.
In this method we endeavor simply to find what propositions
have already been proved that will lead to the proof of the
proposition that is before us. This method was used in all the
exercises on pages 28, 31, and 33.
(2) By superposition. In Props. II and III we placed one figure
on another and then, by synthetic reasoning, showed them to
be identically equal. Such proof is known as a proof by super
position. Superposition means " placing on," and one figure is
said to be superposed on the other.
In Prop. V a special kind of proof by superposition was
employed, in which we superpose a figure on its exact dupli
cate. This special method is rarely used, but in this proposi
tion it materially simplifies the proof.
79. Converse Propositions. If two propositions are so related
that what is given in each is what is to be proved in the other,
each proposition is called the converse of the other.
E.g. in Prop. IV we have given
AC = BC, to prove that ZA = ZB.
In Prop. V we have given
ZA = ZB, to prove that AC = BC.
Hence Prop. V is tlie converse of Prop. IV, and Prop. IV is the con
verse of Prop. V.
Not all converses are true, and hence we have to prove any
given converse.
E.g. tlie converse of the statement ''Two right angles are two equal
angles" is "Two equal angles are two right angles," and this statement
is evidently false.
36 BOOK I. pla:^^e geometry
Proposition VI. Theorem
80. Tioo triangles are congruent if the three sides of the
one are equal respectively to the three sides of the other.
Given the triangles ABC and A'B'C\ with AB equal to A^B^
AC equal to A'C, and BC equal to B'C
To prove that A ABC is congruent to AA'B'C.
Proof. Let AB and A'B' be the greatest of the sides of the A.
Place A A'B'C next to A ABC so that A' shall fall on A,
the side A'B' shall fall along AB, and the vertex C' shall be
opposite the vertex C. Post. 5
Then B' will fall on B.
{For A'B'' is given equal to AB.)
Draw CC.
Since AC = AC', Given
.•.ZACC' = ZCCA. §74
Since BC=BC', Given
.'.ZC'CB = ZBC'C. §74
.. Z .1 CC"+ Z C'CB = Z CCA + Z BC'C. Ax. 1
Hence ZylC5 = Z5C"^. Ax. 11
{For Z A CB is made up of ZACC and Z C'CB, and A BC'A is made
up of Z CCA and Z BC'C.)
.. A ABC is congruent to A ABC'. § 68
.*. A.4J5C is congruent to A.l'^'C', by Ax. 9. q.e.d.
TRIANGLES
37
EXERCISE 8
1. Prove that a line from the vertex to the midpoint of
the base of an isosceles triangle cuts the triangle into two
congruent triangles.
2. Three iron rods are hinged at the ex
tremities, as shown in this figure. Is the
figure rigid ? Why ?
3. Four iron rods are hinged, as shown in
this figure. Is the figure rigid ? If not, where
would you put in the fifth rod to make it rigid ?
Prove that this would accomplish the result.
4. If two isosceles triangles are constructed on opposite
sides of the same base, prove by Prop. VI and § 58 that
the line through the vertices bisects the
vertical angles.
5. In this figure AB = AD and CB = CD. ^'
Prove that AC bisects Z BAD and Z DCB.
6. In § 31, Ex. 8, it was shown how to bisect
an angle, this being the figure used. Draw PX
and PY, and prove by Prop. VI that PO bi
sects ZAOB.
7. In a triangle ABC it is known that A C = BC.
It A A and Z B are both bisected by lines meet
ing at P, prove that AABP is isosceles.
8. In this figure it is known that Am = A n.
Prove that ^C = ^C.
9. From the vertices A and B of an equilateral triangle
lines are drawn to the midpoints of the opposite c
sides. Prove that these two lines are equal.
In A ABQ and BAP show that the conditions of
congruence as stated in Prop. II are fulfilled.
38 BOOK L PLANE GEOMETRY
Proposition VII. Theorem
81. The sum of tivo lines from a given point to the
extremities of a given line is greater than the sum of
tivo other lines similarly drawn, hut included hy them.
B
Given CA and C5, two lines drawn from the point C to the
extremities of the line AB^ and PA and PB two lines similarly
drawn, but included by CA and CB.
To prove that CA\CB>PA} PB.
Proof. Produce AP to meet the line CB at Q. Post. 2
Then CA \ CQ>PA ^ PQ. Post. 3
{A straight line is the shortest path between two points.)
Likewise .BQ\PQ>PB. Post. 3
Add these inequalities, and we have
CAJrCQ + BQ + PQ>PA+PQ^PB. Ax. 7
{If unequals are added to unequals in the same order, the sums are unequal
in the same order.)
Substituting for CQ + BQ its equal CB, we have
CA\CBi PQ >PA +PQ + PB. Ax. 9
{A quantity may he substituted for its equal in an equation or in an
inequality.)
Taking PQ from each side of the inequality, we have
CA^CB>PA\ PB, by Ax. 6. q. e. d.
TEIANGLES 39
Proposition VIII. Theorem
82. Only one perpendicular can he drawn to a given
line from a given external point.
Given a line XF, P an external point, PO a perpendicular to XY
from P, and PZ any other line from P to XY.
To prove that PZ is not ± to XY.
Proof. Produce PO to P', making OP' equal to PO. Post. 2
Draw P'Z. Post. 1
By construction POP' is a straight line.
.'. PZP' is not a straight line. Post. 1
Hence Z P'ZP is not a straight angle. § 33
Since A POZ and ZOP' are rt. A, § 27
.'.ZP0Z = ZZ0P'. §56
Furthermore PO = OP', Hyp.
and OZ — OZ. Iden.
.. A OPZ is congruent to AOP'Z, § 68
( Two A are congruent if two sides and the included Z of the one are
equal respectively to two sides and the included Z of the other.)
and Z OZP = Z P'ZO. § 67
.*. Z OZP, the half of Z P'ZP, is not a right angle. § 34
.. PZ is not ± to Xr, by § 27. Q.e.d.
40 BOOK I. PLANE GEOMETEY
Proposition IX. Theorem
83. Tioo lines draion from a point in a perpendicu
lar to a given line, cutting off on the given line equal
segments from the foot of the perpendicular, are equal
and make equal angles ivith the perpendicular.
A o B ^
Given PO perpendicular to XF, and PA and PB two lines cutting
off on XY equal segments OA and OB from 0.
To prove that PA = PB,
and ZAPO = ZOPB.
Proof. In the A A OP and BOP,
Z POA and Z BOP are rt. A. § 27
{For PO is given ± to XY.)
.'.ZP0A=ZB0P. §56
{All right A are equal.)
Also OA = OB, Given
and PO = PO. Men.
{That is, PO is common to the two ^.)
..A A OP is congruent to A BOP. § 68
( Two A are congruent if two sides and the included Z of the one are equal
respectively to two sides and the included Z of the other.)
.. PA = PB,
and AAPO = Z OPB. § 67
{Corresponding parts of congruent figures are equal.) Q.E.D.
TRIANGLES
Proposition X. Theorem
41
84. Of two lines draion from a point in a perpen
dicular to a given line, cutting off on the given line
unequal segments from the foot of the perpendicidar,
the more remote is the greater.
Given PO perpendicular to XY, PA and PC two lines drawn
from P to XY, and OA greater than OC.
To prove that PA>PC.
Proof. Take OB equal to OC, and draw PB.
Then PB = PC. § 83
Produce PO to P', making OP' = PO, and draw P'A and P'B.
Then PA = P'A and PB = P'B. § 83
But PA + P'A >PB + P'B. § 81
. • . 2 PA >2PB and PA > PB. Axs. 9 and 6
.'.PA >PC, by Ax. 9. q.e.d.
85. Corollary. Onl^ two equal obliques can be drawn from
a given point to a given line, and these cut off equal segments
from the foot of the perpendicular.
Of two unequal lines from a point to a line, the greater cuts
off the greater segment from the foot of the perpendicular.
For PB = PC, but PB cannot equal PA (§ 84). The segments OB
and OC are equal, for otherwise PB could not equal PC
42
BOOK I. PLAKE GEOMETRY
Proposition XI. Theorem
86. The perpendicular is the shortest line that can he
drawn to a given line from a given external point.
Given a line XF, P an external point, PO the peri)endicular, and
PZ any other line drawn from P to XY.
To prove that PO < PZ.
Proof. Produce PO to P', making OP' = PO) and draw P'Z.
Then PZ = P'Z. § 83
{Two lines drawn from a point in a 1. to a given line, cutting off on the
given line equal segments from the foot of the ±, are equal.)
.•.PZ + P'Z = 2 PZ. Ax. 1
Furthermore PO + P'O = 2 PO. Ax. 1
But PO + P'O <PZ\ P'Z. Post. 3
.'.2PO<2PZ. Ax. 9
.'. PO<PZ, by Ax. 6. q.e.d.
87. Hypotenuse. The side opposite the right angle in a right
triangle is called the hypotenuse.
The other two sides of a right triangle are usually called the sides.
88. Distance. The length of the straight line from one point
to another is called the distance between the points.
The length of the perpendicular from an external point to a
line is called the distance from the point to the line.
TRIANGLES 43
Proposition XII. Theorem
89. Tico right triangles are congruent if the hypote
nuse and a side of the one are equal respectively to the
hypotenuse and a side of the other.
A
/
Given the right triangles ABC and A'5'C', with the h3rpotenuse
AC equal to the hypotenuse A^C\ and with BC equal to 5'C'.
To prove that A ABC is congruent to AA'B'C.
Proof. Place A ABC next to A.4'^'C', so that BC shall fall
along B'C, B shall fall on B'j and A and A ' shall fall on opposite
sides of 5'C'.
Then
and
Since
C will fall on C,
{For BC is given equal to B'C\)
BA will fall along A 'B' produced. ,
{For A CBA + Z A'B'C'= a st. Z.)
AC' = A'C\
Post. 5
§34
.•.AB' = A'B'. §85
.. A ABC is congruent to A A^B^C. § 80
( Two A are congruent if the three sides of the one are equal respectively
to the three sides of the other.) Q. E. D.
90. Corollary. Two right triangles are congruent if any
two sides of the one are equal respectively to the corresponding
two sides of the other.
44
BOOK I. PLANE GEOMETRY
EXERCISE 9
1. ABCD is a square and Mis the midpoint of AB. With
ikf as a center an arc is drawn, cutting 5C at P and AD Sit Q.
Prove that A MBP is congruent to A MA Q, and d c
write the general statement of this theorem
without using letters as is done here.
This would read, "If an arc is drawn, with the mid
point of one side of a square as a center, cutting the ^ ^
sides perpendicular to that side, then the triangles cut off by," etc.
2. Draw a figure similar to that of Ex. 1, but take a radius
such that the arc cuts BC produced at a point above C, and
AD above D. Then prove that A MBP is congruent to A MA Q.
3. Prove that if from the point P the perpendiculars PM,
PN to the sides of an angle A OB are equal, the
point P lies on the bisector of the angle A OB. JVx
Write the general statement of this theorem
without using letters as is done here. ^ ^
4. Prove that if the perpendiculars from the midpoint M of
the base AB oi a, triangle ^i^C to the sides of the
triangle are equal, then AA=A B. What then
follows as to the sides A C and BC ? Write the gen
eral statement of this theorem without referring
to a special figure.
5. Prove that if the perpendiculars from the extremities of
the base of a triangle to the other two sides are equal, the
triangle is isosceles.
6. Suppose 0Y1.0X. With as a center an arc is drawn
cutting OX at A and OF at 5. Then with A
as a center an arc is drawn cutting OF at P,
and with 5 as a center and the same radius
an arc is drawn cutting OX at Q. Prove
that OP = OQ.
What triangles are congruent by Prop. XII ?
TRIANGLES 45
Proposition XIII. Theorem
91. Tico right triaiigles are congruent if the hypotenuse
and an adjacent angle of the one are equal respectively
to the hyjjotenuse and an adjacent angle of the other,
c
Given the right triangles ABC, A'B'C\ with the hypotenuse AC
equal to the hypotenuse A'C, and with angle A equal to angle A'.
To prove that A ABC is congruent to AA'B'C.
Proof. Place A ABC upon A A'B'C so that A shall fall upon
A' SiudAC shall fall along A 'C. Post. 5
Then C will fall onC,
{For AC is given equal to A'C\)
and AB will lie along A'B'.
{For ZA is given equal to Z A'.)
Then because C falls on C",
and AB and B^ are rt. A, Given
{Since the A are given as ri. A.)
.. CB will coincide with C'B'. § 82
{Only one perpendicular can be drawn to a given line from a
given external point.)
.'. A ABC is congruent to A A'B'C^. § 66
{If two figures can he made to coincide in all their parts,
they are said to he congruent.) Q.E.D.
46 BOOK I. PLANE GEOMETRY
Proposition XIV. Theorem
92. TiDO lines in the same plane perpendicular to the
same line cannot meet hoiveve?' far they are produced.
X
Given the lines AB and CD perpendicular to XY at A and C
respectively.
To prove that AB and CD cannot meet however far they
are produced.
Proof, li AB and CD can meet if sufficiently produced, we
shall have two perpendicular lines from their point of meeting
to the same line.
But this is impossible. § 82
.'. AB and CD cannot meet. q.e.d.
93. Parallel Lines. Lines that lie in the same plane and
cannot meet however far produced are called parallel lines.
94. Postulate of Parallels. Through a given point only one
line can he drawn parallel to a given line.
As always in such cases the word line means straight line.
95. Corollary 1. Two lines iii the same plane perpen
dicular to the same line are parallel.
96. Corollary 2. Two lines in the same plane parallel to
a third line are parallel to each other.
For if they could meet, we should have two lines through a point
parallel to a line. Why is this impossible ?
PARALLEL LINES
47
Proposition XV. Theorem
97. If a line is j^erpendicular to one of tivo parallel
lines, it is perpendicular to the other also.
o
P
^"^
Given AB and Ci>, two parallel lines, with XY perpendicular to
AB and cutting CD at P.
To prove that XY is ± to CD.
Proof. Suppose MN drawn through P _L to XY.
Then MN is II to AB. § 95
But CD is 11 to AB. Given
.*. CD and MN must coincide. § 94
But XF is _L to MN. Hyp.
.'. AT is X to CD. Q.B.D.
98. Transversal. A line that cuts two or more lines is called
a transversal of those lines.
99 . Angles made by a Transversal.
If XY cuts AB and CD, the angles
a, d, g, f are called interior angles ;
h, c, h, e are called exterior angles.
The angles d and /, and a and g,
are called alternateinterior angles ; c
the angles b and h, and c and e, are
called alternateexterior angles.
The angles b and/, c and g, e and a, h and d, are called exterior
interior angles.
48 BOOK I. PLAN^E GEOMETRY
Proposition XVI. Theorem
100. If tivo parallel lines are cut hy a transversal, the
alternateinterior angles are equal.
X
Given AB and CD, two parallel lines cut by the transversal XY
in the points P and Q respectively.
To prove that ZAFQ = ZDQP.
Proof. Through 0, the midpoint of PQ, suppose MN drawn
_L to CD.
Then MN is likewise ±to AB. § 97
{A line ± to one of two \\s is 1. to the other.)
Now A PMO and QNO are rt. A. § 63
{Since A OMP and ONQ are rt. A.)
But Z POM = Z QON, § 60
(If two lines intersect, the vertical A are equal.)
and OP=OQ. Hyp.
{For was taken as the midpoint of PQ.)
.'.A PMO is congruent to A QNO. § 91
( Two right A are congruent if the hypotenuse and an adjacent Z of the one
are equal respectively to the hypotenuse and an adjacent Z of the other.)
.'.ZAPQ = ZDQP. §67
{Corresponding parts of congruent figures are equal.) Q. E. D.
PAKALLEL LINES 49
Proposition XVII. Theorem
101. When two lines in the same plane are cut hy a
transversal, if the alternateinterior angles are equal,
the two lines are parallel.
Y
Given the lines AB and CD cut by the transversal XY in the
points P and Q respectively, so as to make the angles APQ and
DQP equal.
To prove that AB is II to CD,
Proof. Since we do not know that AB is II to CD, let us
suppose MN drawn through P II to CD.
We shall then prove that AB coincides with MN.
Now Z MPQ = Z DQP. § 100
{If two II lines are cut by a transversal, the alt.int. A are equal.)
But Z APQ =:^^ DQP. Given
.'.ZAPQ=:Z.MPQ. Ax. 8
{Quantities that are equal to the same quantity are equal to each other.)
.'. AB and MN must coincide. § 23
(De/. of equal angles.)
But MN is II to CD. Hyp.
{For MN was drawn II to CD.)
.'. AB, which coincides with MN, is II to CD. q.e.d.
This proposition is the converse of Prop. XVI, as defined in § 79.
50 BOOK I. PLANE GEOMETEY
Proposition XVIII. Theorem
102. If tivo parallel lines are cut hy a transversal, the
exteriorinterior angles are equal.
Y
Given AB and CD, two parallel lines, cut by the transversal XY
in the points P and Q respectively.
To prove that Z BPX =ADQX.
Proof. Z BPX = Z.APQ. § 60
ZAPQ = ZDQX. §100
.'. ZBPX = ZDQX, by Ax. 8. q.e.d.
103. Corollary 1. When two lines are cut hy a transversal,
if the exteriorinterior angles are equal, the lines are parallel.
The proofs of §§ 103 and 105 are similar to that of § 101.
104. Corollary 2. If two parallel lines are cut hy a trans
versal, the two interior angles on the same side of the trans
versal are supplementary.
105. Corollary 3. When two lines are cut hy a transversal,
if two interior angles on the same side of the transversal are
supplementary, the lines are parallel.
106. Corollary 4. If two parallel liries are cut hy a trans
versal, the alternateexterior angles are equal.
TEIANGLES 51
Proposition XIX. Theorem
107. The smn of the three angles of a triangle is equal
to two right angles.
c
A
Given the triangle ABC.
To prove that AA\A B\ZC=2 H. A.
Proof. Suppose BY drawn II to AC, and produce AB to X
Then Z XB Y\Z.YBC + Z. CBA = 2 rt. A. § 34
{For a St. Z equals 2 rt. A.)
But AA=AXBY, §102
and AC = ZYBC. §100
.'.ZA+AB\AC = 2Yt.A, by Ax. 9. q.e.d.
108. Corollary 1. Tf two triangles have two angles of the
one equal to two angles of the other, the third angles are equal.
109. Corollary 2. In a triangle there can he hut one right
angle or one obtuse angle.
110. Exterior Angle. The angle included by one side of a
figure and an adjacent side produced is called an exterior angle.
In the above figure A XBC is an exterior angle, and A A and C are
called the opposite interior angles.
111. Corollary 3. An exterior angle of a triangle is equal
to the sum of the two opposite interior angles, and is therefore
greater than either of them.
62
BOOK I. PLANE GEOMETRY
EXERCISE 10
1. Show that if we place a draftsman's
triangle against a ruler and draw A C, and
move the triangle along as shown in the
figure and draw A'C, then ^C is II to A'C. [
2. In the next figure x = 60°. How many de
grees in each of the other seven angles ?
3. In the next figure representing two pairs
of parallel lines certain angles are equal. State
these equalities in this form : a = c = g = e =
= •■', and give the reason in each case.
4. In the figure of Ex. 3 state ten pairs
of nonadjacent angles that are supplementary.
Thus : a\h= 180° and cZ + e == 180°.
5. In the triangle ABC, AC =BC and DE is
drawn parallel to AB. Prove that CD = CE.
Write a general statement of the theorem.
6. In the next figure A B is parallel to CD, and
/APQ is half of Z.QPB. How many degrees in
the various angles ?
7. If Z YQD = 135°, how many degrees in the
various angles ?
8. Let ZDQP = x and Z.YQD = y. Then if
y — x = 100°, find the value of x and y.
9. Let ZCQY = x and ZXPA=y.
the value of x and y.
10. In the next figure x = 72° and x = ^y. It is required to
know if the lines are parallel, and why.
11. In the figure of Ex. 10 suppose x = 73° and
y — x = 32°. It is required to know if the lines
are parallel, and why.
Then if x = ^y, find
TRIANGLES 53
The three angles of a triangle are x, g, and z. Find the
value of 0, given the values of x and y as follows :
12. X = 10°, y = 30°. 11. x = 37°, y = 48°.
13. a; = 20°, 2/ = 20°. 18. ic = 63°, y = 29°.
14. X = 75°, y = 50°. 19. a; =75° 29', y = 68° 41'.
15. X = 38°, y = 76°. 20. x = 82° 33', y = 75° 48'.
16. X = 49°, y = 92°. 21. ic = 69° 58', y = 82° 49'.
22. In a certain right triangle one angle is 37°. What is the
size of the other acute angle ?
23. In a certain right triangle one angle is 36° 41'. What is
the size of the other acute angle ?
24. In a certain right triangle one angle is 29° 48' 56".
What is the size of the other acute angle ?
25. In a certain right triangle one acute angle is two thirds
of the other. How many degrees are there in each ?
26. In a certain right triangle one acute angle is twice as
large as the other. How many degrees are there in each ?
27. In a certain right triangle the acute angles are 2 x and
5 X. Find the value of x and the size of each angle.
28. In a certain triangle one angle is twice as large as
another and three times as large as the third. How many
degrees are there in each ?
29. In a certain isosceles triangle one angle is twice another
angle. How many degrees in each of the three angles ?
30. In this figure what single angle equals
a \ G? To the sum of what angles is q equal ?
also r ? From these relations find the number y^ ^
of degrees in ^9 + 5' + r.
31. Prove Prop. XIX by first drawing a parallel to AB
through C, instead of drawing BY.
54 BOOK L PLANE GEOMETRY
Proposition XX. Theorem
112. The sum of any two sides of a triangle is greater
than the third side, and the difference hetiveen any tivo
sides is less than the third side.
A B
Given the triangle ABC^ with AB the greatest side.
To prove that BC + CA >AB, and AB BC< CA.
Proof. BC + CA >AB. Post. 3
{A straight line is the shortest path between two points.)
Since BC]CA>AB,
.■.CA>ABBC; Ax. 6
or, AB — BC<CA. q.e.d.
EXERCISE 11
State in what cases it is possible to form triangles with rods
of the following lengths, and give the reason :
1. 2 in., 3 in., 4 in. 4. 7 in., 10 in., 20 in.
2. 3 in., 4 in., 7 in. 5. 8 in., 9^ in., 18 in.
3. 6 in., 7 in., 9 in. 6. 9f in., lOi in., 12i in.
7. In this figure prove that AB \BC >AD ]DC.
Why is DB+BODC?
What is the result of adding AD to these unequals ?
8. How many degrees are there in each
angle of an equiangular triangle ? Prove it. ^ ^ ^
TRIANGLES 55
Proposition XXI. Theorem
113. If two sides of a triangle are unequal , the angles
opjjosite these sides are unequal, and the angle opposite
the greater side is the greater.
A B
Given the triangle ABC^ with BC greater than CA.
To prove that ZBAOZB.
Proof. On CB suppose CX taken equal to CA.
Draw AX. Post. 1
Then A AXC is isosceles. § 62
Then Z CXA = Z XA C. § 74
(In an isosceles A the A opposite the equal sides are equal.)
But Z CXA >ZB. § 111
{An exterior Z of a A is greater than either opposite interior Z.)
Also ZBAOZ XA C. Ax. 11
{For ZXAC isa part of Z BA C.)
Substituting in this inequality for Z XAC its equal, Z CXA,
we have the inequality
Z BA C>Z CXA . Ax. 9
Since
Z^^OZCA'.l,
and
Z.CXA>ZB,
.'.ZBAOZB.
Ax. 10
{If the first of three quantities is greater than the second, and the second is
greater than the third, then the first is greater than the third.) Q. E. D.
56 BOOK T. PLANE GEOMETRY
Proposition XXII. Theorem
114. If two angles of a triangle are unequal, the sides
opposite these angles are unequal, and the side opposite
the greater angle is the greater.
c
Given the triangle ABC^ with the angle A greater than the angle B.
To prove that BOCA.
Proof. Now BC is either equal to CA, or less than CA, or
greater than CA.
But ii BC were equal to CA,
then the Z A would be equal to the Z.B. § 74
{For they would he A opposite equal sides.)
And if CA were greater than BC,
then the Z B would be greater than the Z.A. § 113
But if CA is not greater than BC, this is only another way
of saying that BC is not less than CA.
We have, therefore, two conclusions to be considered,
ZA=ZB,
and ZA<ZB.
Both these conclusions are contrary to the given fact that
the Z ^ is greater than the Z B.
Since BC cannot be equal to CA or less than CA without
violating the given condition, .'. BOCA. q.e.d.
This proposition is the converse of Prop. XXL
TRIANGLES
Proposition XXIII. Theorem
57
115. If tivo triangles have two sides of the one equal
respectively to two sides of the other, hut the included
angle of the first triangle greater than the included
angle of the second, then the third side of the first is
greater than the third side of the second.
Y Y
Given the triangles ABC and XYZ, with CA equal to ZX and
BC equal to FZ, but with the angle C greater than the angle Z.
To 2)rove that AB >XY.
Proof. Place the A so that Z coincides with C and ZX falls
along CA. Then .A' falls on A, since ZX is given equal to CA,
and Z Y falls within Z A CB, since AC is given greater than Z Z.
Suppose CP drawn to bisect the Z YCB, and draw YP.
Then since
CP = CP,
CY = CB,
Iden.
Given
d
Zycp = Zpcb
}
Hyp.
.
'. A PYC is congruent
to A PBC.
§68
.•.PY=PB.
§67
Now
AP{PY>AY.
.•.AP^PB>AY.
.\AB>AY.
Post. 3
Ax. 9
Ax. 11
.\AB>XY,
by
Ax.
9.
Q.E.D.
68 BOOK I. PLANE GEOMETRY
Proposition XXIV. Theorem
116. If tivo triangles have two sides of the one equal
respectively to tivo sides of the other ^ hut the third side
of the first triangle greater than the third side of the
second, then the angle opposite the third side of the first
is greater than the angle opposite the third side of
the second.
z
A B X Y
Given the triangles ABC and XYZ^ with CA equal to ZX and BC
equal to FZ, but with AB greater than XY.
To prove that the Z. C is greater than the AZ.
Proof. Kow the Z C is either equal to the Z Z, or less than
the Z Z, or greater than the Z Z.
But if the Z C were equal to the Z Z,
then the A ^5C would be congruent to the A XFZ, § 68
{^or it would have two sides and the included Z of the one equal respectively
to two sides and the included Z of the other.)
and AB would be equal to XY. § 67
And if the Z C were less than the Z Z,
then AB would be less than XY. § 115
Both these conclusions are contrary to the given fact that
AB is greater than XY.
.'. AOAZ. Q.E.D.
This proposition is the converse of Prop. XXIII.
QUADRILATERALS 59
117. Quadrilateral. A portion of a plane bounded by four
straight lines is called a quadrllateraL
118. Kinds of Quadrilaterals. A quadrilateral may be
a trapezoid, having two sides parallel ;
2^ parallelogram, having the opposite sides parallel.
If the^nonparallel sides are equal, a trapezoid is called isosceles.
A quadrilateral with no two sides parallel is called a trapezium.
Trapezoid Parallelogram Tiai)eziuin
119. Kinds of Parallelograms. A parallelogram may be
a rectangle, having its angles all right angles ;
a rhombus, having its sides all equal.
A parallelogram with all its angles oblique is called a rhomboid.
Rectangle Rhombus Rhomboid
120. Base. The side upon which a hgure is supposed to
rest is called the base.
If a quadrilateral has a side parallel to the base, this is called the
wpper base, the other being called the lower base.
In an isosceles triangle the vertex formed by the equal sides is taken
as the vertex of the triangle, and the side opposite this vertex is taken as
the base of the triangle.
121. Altitude. The perpendicular distance between the bases
of a parallelogram or trapezoid is called the altitude.
The perpendicular distance from the vertex of a triangle to
the base is called the altitude of the triangle.
122. Diagonal. The straight line joining two nonconsecutive
vertices of any figure is called a diagonal.
60 BOOK I. PLANE GEOMETRY
Proposition XXV. Theorem:
123. TiDO angles whose sides are parallel each to each
are either equal or supplementary.
z
Given the angle AOB and the lines WY and XZ parallel to the
sides and intersecting at P, the figure being lettered as shown.
To prove that /.p = /LO, and that Z.p' is supplementary
to ZO.
Proof. Let OA meet XZ at M. Then in the figure
ZO = Z 7/1, and Zp = Z m. § 102
{If two II lines are cut by a transversal, the ext.int. A are equal.)
.■.Zp = Z.O. Ax. 8
Also ZL])' is the supplement of Zp. § 42
.*. Zp' is supplementary to Z 0, by § 58. q.e.d.
If the sides of two angles are parallel each to each, under what
circumstances are the angles equal, and under what circumstances are
they supplementary ?
124. Corollary. The opposite angles of a parallelogram
are equal, and any two consecutive angles are supplementary.
Draw the figure and explain how it is known that any angle is the
supplement of its consecutive angle. If two opposite angles are supple
ments of the same angle, show that § 58 applies.
QUADRILATERALS 61
Proposition XXVL Theorem
125. The opposite sides of a i^arallelogram are equal.
A
Given the parallelogram ABCD.
To prove that BC = AD, and AB = DC.
Proof. Draw the diagoDal A C.
In the A ABC and CDA,
AC = AC, Iden.
ZBAC == ZDCA,
and ZACB = Z CA D. § 100
.. A ABC is congruent to A CDA. § 72
.. BC = AD, and AB = DC, by § 67. Oe.d.
126. Corollary 1. A diagonal divides a parallelogram into
two congruent triangles.
Upon what theorem does this depend ?
127. Corollary 2. Segments of parallel lines cut off hy
parallel lines are equal.
How does this follow from the proposition ?
128. Corollary 3. Two parallellines are everywhere equally
distant from each other. A B
\i AB and CD are parallel, what can
be said of Js dropped from any points in
AB to CD (§ 127) ? Hence what may
be said of all points in AB with respect to their distance from CD ?
62 BOOK L PLANE GEOMETRY
Proposition XXVII. Theorem
129. If the opposite sides of a quadrilateral are equal,
the figure is a parallelogram.
Given the quadrilateral ABCD^ having BC equal to AD^ and
AB equal to DC.
To prove that the quadrilateral ABCD is a parallelogram.
Proof. Draw the diagonal A C.
In the A ABC ^nd'CDA,
BC = AD, Given
AB = DC, Given
and AC = AC. Iden.
.. A ABC is congruent to A CDA. § 80
( Two A are congruent if the three sides of the one are equal respectively
to the three sides of the other.)
.. ZBAC==ZDCA,
and ZACB = ZCAD. §67
.. AB is II to DC,
and BC is II to AD. § 101
{When two lines in the same plane are cut by a transversal, if the
alt. int. A are equal, the two lines are II.)
..the quadrilateral ABCD is a O, by § 118. Q.e.d.
This proposition is the converse of Prop. XXVI.
QUADEILATERALS 63
Proposition XXVIII. Theorem
130. If two sides of a quadrilateral are equal and
parallel, then the other two sides are equal and par
allel, and the figure is a p)arallelogram.
Given the quadrilateral ABCD^ having AB equal and parallel
to DC.
To prove that the quadrilateral ABCD is a parallelogram.
Proof. Draw the diagonal A C.
In the A ABC and CDA,
AC = AC, Iden.
AB = DC, Given
and ABAC = A DC A . § 100
{If two II lines are cut by a transversal, the alt.int. A are equal.)
..A ABC is congruent to A CDA. § 68
{Two A are congruent if two sides and the included Z of the one are
equal respectively to two sides and the included Z of the other.)
.\BC = AD,
and ZACB = Z CAD. § 67
.. BC is II to AD. § 101
( When two lines in the same plane are cut by a transversal, if the
alt.int. A are equal, the two lines are II.)
But AB is II to DC. Given
.. the quadrilateral ABCD is a O, by § 118. q.e.d.
64 BOOK I. PLANE GEOMETRY
Proposition XXIX. Theorem
131. The diagonals of a parallelogram bisect each
other.
G
A
B
Given the
parallelogram
ABCD
, with the
diagonals
AC and BD
intersecting
at 0.
To prove
that
A0 =
OC,
and
B0 =
01).
Proof. If we can show that the A ABO is congruent to the
AC DO, or that the ABCO is congruent to the ADAO, the
proposition is evidently proved, since the corresponding sides
of the congruent triangles will be equal.
Now in the A ABO and CDO,
AB = CD, §125
{The opposite sides of a O are equal.)
Z.BAO = Z.DCO,
and Z OBA = Z ODC. § 100
{If two parallel lines are cut by a transversal, the alternateinterior
angles are equal.)
.'. AABOi^ congruent to A CDO. § 72
{Two A are congruent if two A and the included side of the one are
equal respectively to two A and the included side of the other.)
.\AO = OC,
and BO = OD. § 67
{Corresponding parts of congruent A are equal.) Q. E. D.
QUADRILATERALS 65
Proposition XXX. Theorem
132. Two j^ciraUelogranis are congruent if tivo sides
and the included angle of the one are equal respectively
to tivo sides and the included angle of the other.
A B A' B'
Given the parallelograms ABCD and A'B'C^D', with AB equal
to A'B', AD to A'Z)', and angle A to angle A'.
To prove that the UJ are congruent.
Proof. Place the EJABCD upon the CJA'B'C'D' so that AB
shall fall upon and coincide with its equal, A'B'. Post. 5
Then AD will fall along A'D\
(For Z A is given equal to Z A'.)
and D will fall on D'.
{For AD is given equal to A^D^.)
Now DC and D'C are both II to A 'B' and are dl'awn through D'.
.'.DC will fall along D'C. ' § 94
(Through a given point only one line can be drawn W to a given line.)
Also BC and B'C' are both II to A 'D' and are drawn through B'.
 .'. BC will fall along B'C. § 94
.. C will fall on C. § 55
.'. the two [U coincide and are congruent, by § 66. q.e.d.
133. Corollary. Two rectangles having equal bases and
equal altitudes are congruent.
How is this shown to be a special case under the above proposition ?
What sides are equal, and what inchided angles are equal ?
66 BOOK I. PLANE GEOMETRY
Proposition XXXI. Theorem
134. If three or more parallels intercept equal, seg
ments on one transversal, they intercept equal segme7its
on every transversal.
Given the parallels AB^ CD^ EF, GH, intercepting equal segments
JBD, DF, FH on the transversal BH^ and intercepting the segments
AC J CEy EG on another transversal.
To prove that AC = CE = EG.
Proof. Suppose AP, CQ, and ER drawn II to BH.
A APC, CQE, ERG = A BDC, DFE, FHG respectively. § 102
But A BDC, DFE, FHG are equal. § 102
.. A APC, CQE, ERG are equal. Ax. 8
AP, CQ, ER are parallel. § 96
Also A CAP, ECQ, GER are equal. § 102
Now AP = BD, CQ = DF, ER = FH. § 127
{Segments of parallels cut off by parallels are equal.)
But BD = DF=FH. Given
.'. AP = CQ = ER. Ax. 8
.'. A CPA, EQC, and GRE are congruent. § 72
.'.AC = CE = EG, by § 67. Q.e.d.
QUADRILATERALS
67
135. Corollary 1. If a line is parallel to one side of a tri
angle and bisects another side, it bisects the third side also.
Let BE be II to BC and bisect AB. Suppose a line is drawn through
A II to BC. Then how do we know this line to
be II to DE ? Since it is given that the three
lis intercept equal segments on the transversal
AB, what can we say of the intercepted seg
ments on AC ? What can we then say that DE
does to AC?
Write the proof of this corollary in full.
136. Corollary 2. The line ivhich joins the midpoints
of two sides of a triangle is parallel to the third side, and is
equal to half the third side.
A line DE drawn through the midpoint of AB, II to BC, divides AC
in what way (§ 135) ? Therefore the line joining the midpoints oi AB
and AC coincides with this parallel and is II to
BC. Also since JPF drawn II to AB bisects AC,
how does it divide BC ? What does this prove
as to the relation of BF, FC, and BC ? Since
BFED is a O (§118), what do we know as to
the equality of DE, BF, and i BC ?
Write the proof of this corollaiy in full.
137. Corollary 3. Tlie line joining the midpoints of the
nonparallel sides of a trapezoid is parallel to the bases and
is equal to half the sum of the bases.
F^^
A
B
Draw the diagonal DB. In the A ABD
join E, the midpoint of AD, to F, the mid
point of DB. Then, by § 136, what relations
exist between EF and AB? In the A DBC
join F to G, the midpoint of BC. Then what
relations exist between FG and DC? Since
this relation exists, what relation exists between AB and FG ? But only
one line can be drawn through F W to AB {% 94). Therefore FG is the
prolongation of EF. Hence EFG is parallel to AB and CD, and equal
to ^{ABk DC).
Write the proof of this corollary in full.
68 BOOK I. PLAXE GEOMETRY
138. Polygon. A portion of a plane bounded by a broken
line is called ?^ polygon.
The terms sides, perimeter, angles, vertices, and diagonals are employed
in the usual sense in connection with polygons in general.
139. Polygons classified as to Sides. A polygon is
a triangle, if it has three sides ;
a quadrilateral, if it has four sides ;
a pentagon, if it has five sides ;
a hexagon, if it has six sides.
These names are sufficient for most cases. The next few names in
order are heptagon, octagon, nonagon, decagon, undecagon, dodecagon.
A polygon is equilateral, if all of its sides are equal.
140. Polygons classified as to Angles. A polygon is
eqtdangular, if all of its angles are equal ;
convex, if each of its angles is less than a straight angle ;
concave, if it has an angle greater than a straight angle.
Equilateral Equiangular Hexagon Convex Concave
An angle of a polygon greater than a straight angle is called a reentrant
angle. When the term polygon is used, a convex polygon is understood.
141. Regular Polygon. A polygon that is both equiangular
and equilateral is called a regular polygon.
142. Relation of Two Polygons. Two polygons are
mutually equiangular, if the angles of the one are equal to
the angles of the other respectively, taken in the same order ;
TRutually equilateral, if the sides of the one are equal to the
sides of the other respectively, taken in the same order ;
congruent, if mutually equiangular and mutually equilateral,
since they then can be made to coincide.
POLYGONS
69
Proposition XXXII. Theorem
143. The sum of the interior angles of a polygon is
equal to two right angles, taken as r)iamj times less two
as the figure has sides.
Given the polygon ABCDEFy having n sides.
To ])rove that the sum of the interior A =(n— 2^2 rt, A.
Proof. From A draw the diagonals AC, AD, AE.
The sum of the A of the A is equal to the sum of the A of
the polygon. Ax. 11
Now there are (n — 2) A.
{For there is one A for each side except the two sides adjacent to A.)
The sum of the A of each A = 2 rt. ^. § 107
.*. the sum of the A of the (n — 2) A, that is, the sum of the
A of the polygon, is equal to (n — 2)2 rt. A, by Ax. 3. q.e.d.
144. Corollary 1. The sum of the angles of a quadrilateral
equals four right angles ; and if the angles are all equal, each
is a right angle.
145. Corollary 2. Uach angle of a regular polygon of n
sides is equal to — ' right angles.
n
70 BOOK I. PLANE GEOMETRY
EXERCISE 12
1. What is the sum of the angles of («) a pentagon ? (h) a
hexagon ? (c) a heptagon ? (d) an octagon ? (6) a decagon ?
(/) a dodecagon? (^) a polygon of 24 sides?
2. What is the size of each angle of (<() a regular pentagon ?
(^) a regular hexagon ? (c) a regular octagon ? (c?) a regular
decagon ? (e) a regular polygon of 32 sides ?
3. How many sides has a regular polygon, each angle of
which is 1 right angles ?
4. How many sides has a regular polygon, each angle of
which is If right angles ?
5. How many sides has a regular polygon, each angle of
which is 108°?
6. How many sides has a regular polygon, each angle of
which is 140°?
7. How many sides has a regular polygon, each angle of
which is 156°?
8. Four of the angles of a pentagon are 120°, 80°, 90°, and
100° respectively. Find the fifth angle.
9. Five of the angles of a hexagon are 100°, 120°, 130°, 150°,
and 90° respectively. Find the sixth angle.
10. The angles of a quadrilateral are x, 2x, 2x, and 3x.
How many degrees are there in each ?
11. The angles of a quadrilateral are so related that the sec
ond is twice the first, the third three times the first, and the
fourth four times the first. How many degrees in each ?
12. The angles of a hexagon are x, 3x, 3x, 2x, 2x, and x.
How many degrees are there in each ?
13. The sum of two angles of a triangle is 100° and their
difference is 40°. How many degrees are there in each of the
three angles of the triangle ?
POLYGONS 71
Proposttton XXXTir. Theorem
146. Tlie sum of the exterior angles of a polijcjon,
made by j^'^oducing each of its sides hi succession, is
equal to four right angles.
Given the polygon ABCDE, having its n sides produced in
succession.
To prove that the sum of the exterior A = 4:rt. A.
Proof. Denote the interior A of the polygon by a, b, c, d, e,
and the corresponding exterior A by a', b', c', d', e'.
Then, considering each pair of adjacent angles,
A a j Aa'= a st. A,
and Ab{Ab'= ^st A. §43
( The two adjacent A which one straight line makes with another are
together equal to a straight Z.)
In like manner, each pair of adj. zi = a st. A.
But the polygon has n sides and n angles.
Therefore the sum of the interior and exterior A§ is equal
to n st. A, OT 2n rt. A. Ax. 3
But the sum of the interior A = (n2)2 rt. A § 143
= 2n rt. Zs  4 rt. A.
.*. the sum of the exterior A = 4: rt. A, by Ax. 2. q.e.d.
72 BOOK I. PLANE GEOMETRY
EXERCISE 13
1. An exterior angle of a triangle is 130° and one of the
opposite interior angles is 52°. Find the number of degrees in
each angle of the triangle.
2. Two consecutive angles of a rectangle are bisected by
lines meeting at P. How many degrees in the angle P ?
3. Two angles of an equilateral triangle are bisected by lines
meeting at P. How many degrees in the angle P ?
4. The two base angles of an isosceles triangle are bisected
by lines meeting at P. The vertical angle of the triangle is 30°.
How many degrees in the. angle P?
5. The vertical angle of an isosceles triangle is 40°. This
and one of the base angles are bisected by lines meeting at P.
How many degrees in the angle P ?
6. One exterior angle of a parallelogram is one eighth of the
sum of the four exterior angles. How many degrees in each
angle of the parallelogram ?
7. How many degrees in each exterior angle of a regular
hexagon ? of a regular octagon ?
8. In a right triangle one acute angle is twice the other.
How many degrees in each exterior angle of the triangle ?
9. Make out a table showing the number of degrees in each
interior angle and each exterior angle of regular polygons of
three, four, five, • • • , ten sides.
10. If the diagonals of a quadrilateral bisect each other, the
figure is a parallelogram.
11. In this parallelogram ABCD, AP =
CR, and BQ = DS. Prove that PQRS is ^
also a parallelogram. A F b
12. If the midpoints of the sides of a parallelogram are
connected in order, the resulting figure is also a parallelogram.
LOCI OF POINTS 73
147. Locus. The path of a point that moves in accordance
with certain given geometric conditions is called the locus of
the point.
Thus, considering only figures in a plane, a .
point at a given distance from a given line of ^
indefinite length is evidently in one of tw^o lines
parallel to the given line and at the given distance from it. Thus, if ^jB
is the given line and d the given distance, the locus is evidently the
pair of parallel lines XY and X'Y\ ^ ^
The locus of a point in a plane at a given distance r / \
from a given point O is evidently the circle described about < r_\
as a center with a radius r. \ ^ I
The plural of locus (a Latin v^^ord meaning "place") is \^ ^/
loci (pronounced losi).
We may think of the locus as the place of all points that satisfy cer
tain given geometric conditions, and speak of the locus of points. Both
expressions, locus of a point and locus of points, are used in mathematics.
EXERCISE 14
State without proof the following loci in a plane :
1. The locus of a point 2 in. from a fixed point O.
2. The locus of the tip of the minute hand of a watch.
3. The locus of the center of the hub of a carriage wheel
moving straight ahead on a level road.
4. The locus of a point 1 in, from each of two parallel lines
that are 2 in. apart.
5. The locus of a point on this page and 1 in. from the edge.
6. The locus of the point of a round lead pencil as it rolls
along a desk.
7. The locus of the tips of a pair of shears as they open,
provided the fulcrum (bolt or screw) remains always fixed in
one position.
8. The locus of the center of a circle that rolls around another
circle, always just touching it.
74 BOOK I. PLANE GEOMETRY
148. Proof of a Locus. To prove that a certain line or group
of lines is the locus of a point that fulhlls a given condition, it
is necessary and sufficient to prove two things :
1. That any point In the sujyposed locus satisfies the condition.
2. That any point outside the supposed locus does not satisfy
the given condition.
For example, if we wish to find the locus of
a point equidistant from these intersecting lines
AB^ CD^ it is not sufficient to prove that any
point on the anglebisector PQ is equidistant from
AB and CD^ because this may be only part of the locus. It is necessary
to prove that no point outside of PQ satisfies the condition. In fact, in
this case there is another line in the locus, the bisector of the Z BOD,
as will be shown in § 152.
149. Perpendicular Bisector. A line that bisects a given line
and is perpendicular to it is called the per2Je7idicida7' bisector of
the line.
EXERCISE 15
Draw the following loci, giving no proofs :
1. The locus of a point \ in. below the base of a given
triangle ABC.
2. The locus of a point ^ in. from a given line AB.
3. The locus of a point 1 in. from a given point O.
4. The locus of a point ^ in. outside the circle described
about a given point O with a radius 1\ in.
5. The locus of a point \ in. within the circle described
about a given point with a radius 1^ in.
6. The locus of a point ^ in. from the circle described about
a given point with a radius 1^ in.
7. The locus of a point ^ in. from each of two given parallel
lines that are 1 in. apart.
LOCI OF POINTS 75
Proposition XXXIV. Theorem
150. The locus of a point equidistant from the extrem
ities of a given line is the perpendicular bisector of
that line. ^
b['\
Given FO, the perpendicular bisector of the line AB.
To prove that YO is the locus of a point equidistant from
A and B.
Proof. Let P be any point in YO, and C any point not in YO.
Draw the lines PA, PB, CA, and CB.
Since . AO = BO, Given
and OP = OP, Iden.
.. rt. A yl OP is congruent to rt. A BOP. § 90
.\PA=PB. §67
Let CA cut the X at D, and draw DB.
Then, as above, DA = DB.
But. CB<CD{DB.  '— Post. 3
.. CB<CD + DA. Ax. 9
.'. CB<CA. Ax. 11
.*. FO is the required locus, by § 148. q.e.d.
151. Corollary. Two points each equidistant from the
extremities of a line determine the perpendicular bisector of
the line.
76
BOOK I. PLANE GEOMETRY
Proposition XXXV. Theorem
152. The locus of a point equidistant from tivo given
intersecting lines is a pair of lines hisecting the angles
formed hy those lines.
Given XX^ and YY^ intersecting at O, 4C the bisector of angle
X'OF, and BD the bisector of angle YOX.
To prove that the pair of lines AC and BD is the locus of
a point equidistant from XX' and YY',
Proof. Let P be any point on A C or BD, and Q any point not on
AC or BD. Let PM and QR be X to XX', PN and QS to YY'.
Since Z MOP = Z PON, Given
and OP = OP, Iden.
.'. rt. A OMP is congruent to rt. A ONP. § 91
.\PM=PN. §67
Let QS cut AO at P'. Draw PT J_ to XX', and draw QT.
Then, as above, P'T= P'S.
But P'r{P'Q>QT, Post. 3
and QT>QR. §86
.. P'T\P'Q>QR. Ax. 10
Substituting, P'5 + ^'Q > QR, or Q^^ > Q/?. Ax. 9
.'. the pair of lines is the required locus, by § 148. q.e.d.
METHODS OF PKOOF 77
153. The Synthetic Method of Proof. The method of proof in
which known truths are put together in order to obtain a new
truth is called the synthetic method.
This is the method used in most of the theorems already given. The
proposition usually suggests some known propositions already proved,
and from these v^^e proceed to the proof required. The exercises on this
page and on pages 78 and 79 may be proved by the synthetic method.
154. Concurrent Lines. If two or more lines pass through the
same point, they are called concurrent lines.
155. Median. A line from any vertex of a triangle to the
midpoint of the opposite side is called a median of the triangle.
EXERCISE 16
1. If two triangles have two sides of the one equal respec
tively to two sides of the other, and the angles opposite two equal
sides equal, the angles opposite the other two equal sides are
equal or supplementary, and if equal the triangles are congruent.
Let ^C = A'C\ BC = B'C\ and /.B = /.B\
Place AA'B'C on A ABC so that B'C shall coincide with BC, and
ZA^ and ZA shall be on the same side of BC.
Since ZB'= ZB, B'A' will fall along what line ? Then A' will fall at
A or at some other point in BA, as B. If A' falls at A, what do we know
about the congruency of the AA'B'C and ABC ?
If A^ falls at D, what about the congruency of the A A'B'C and BBC ?
Since CB = C'A' = CA, what about the relation of Z ^ to Z CBA ?
Then what about the relation of the A CBA and BBC ?
Then what about the relation of the A A and BBC ?
Draw figures and show that the triangles are congruent :
1. If the given angles B and JB' are both right or both obtuse.
2. If the angles A and A^ are both acute, both right, or both obtuse*
3. If ^C and A'C are not less than BC and B'C respectively.
78
BOOK I. PLANE CxEOMETEY
2. The bisectors of the angles of a triangle are concurrent in
a point equidistant from the sides of the triangle.
The bisectors of two angles, as AD and BE, intersect as at 0.
Why? Now show that is equidistant from AC and ^
AB, also from BC and AB, and hence from AC and
BC. Therefore, where does lie with respect to the ,_g,, p
bisector Ci^? ^^l^^
This point is called the incenter of the triangle. ^ p
3. The perpendicular bisectors of the sides of a triangle are
concurrent in a point equidistant from the vertices.
The ± bisectors of two sides, as QQ^ and RW, intersect as at 0.
Why ? Now show that is equidistant from B
and O, also from C and A, and hence from A
and B. Therefore, where does lie with respect
to the ± bisector PP' ?
This point is called the circumcenter of the
triangle.
4. The perpendiculars from the vertices of a triangle to the
opposite sides are concurrent.
Let the Js be J.Q, BR, and CP. Through A, B, C suppose B'C% C'A\
Bx
lA'
&
and ^'i^' drawn II to CB, AC, and BA respec
tively. Now show that C" J. ==50 = ^B'. In the
same way, what are the midpoints of C'A' and
A'B' ? How does this prove that A Q, BR, and CP J
are the _L bisectors of the sides of the A A''B'C' ?
Proceed as in Ex. 3.
This point is called the orthocenter of the triangle:
5. The medians of a triangle are concurrent in a point two
thirds of the distance from each vertex to the middle of the
opposite side.
Two medians, as ^ Q and CP, meet as at 0. If Y is the midpoint of A 0,
and X of CO, show that YX and PQ are II to ^C and
equal to ^ AC. Then show that AY=YO=OQ, and
CX = XO = OP. Hence any median cuts off on any
other median what part of the distance from the ver
tex to the midpoint of the opposite side ?
This point is called the centroid of the triangle.
METHODS OF PROOF 79
6. The bisectors of two vertical angles are in b
the same straight line. ol^^^
c — ^^ — ^
7. The bisector of one of two vertical angles ^/</
bisects the other. d
8. The bisectors of two supplementary adjacent angles are
perpendicular to each other.
9. The bisectors of the two pairs of vertical angles formed
by two intersecting lines are perpendicular to each other.
10. If the bisectors of two adjacent angles are JY .b
perpendicular to each other, the adjacent angles \/^^
are supplementary. o
11. If an angle is bisected, and if a line is drawn ^ ^^^ ^
through the vertex perpendicular to the bisector,
this line forms equal angles with the sides of the
given angle. o ^
12. The bisector of the vertical angle of an isosceles triangle
bisects the base and is perpendicular to the base.
13. The perpendicular bisector of the base of an isosceles
triangle passes through the vertex and bisects the e
angle at the vertex.
14. If the perpendicular bisector of the base of
a triangle passes through the vertex, the triangle
is isosceles. a d b
15. Any point in the bisector of the vertical angle of an isos
celes triangle is equidistant from the extremities of the base.
16. If two isosceles triangles are on the same base, a line
passing through their vertices is perpendicular to the base
and bisects the base.
17. Two angles whose sides are perpendicular each to each
are either equal or supplementary.
Under what circumstances are the angles equal, and under what
circumstances are they supplementary ?
80 BOOK I. PLANE GEOMETKY
156. The Analytic Method of Proof. The method of proof that
asserts that a proposition under consideration is true if another
proposition is true, and so on, step by step, until a known
truth is reached, is called the analytic method.
This is the method resorted to when we do not see how to start the
ordinary synthetic proof. The exercises on this page and on pages 81
and 82 may be investigated by the analytic method.
EXERCISE 17
1. The midpoint of the hypotenuse of a right triangle is
equidistant from the three vertices.
Given M the midpoint of AC, the hypotenuse of the rt. /\ABC.
To prove that M is equidistant from A, B, and C.
We may reason thus : M is equidistant from A, B, and C if AM= BM.
Why is this the case ?
AM= BM if the J. MN cuts A ABM into two
congruent A. ]\]
A ANM is congruent to A BNM ii AN= NB.
But AN does equal NB (§ 135), because MN
is II to CB, and AM = MC.
Therefore the proposition is true.
We may now, in writing our proof, begin with this last step and work
backwards, as in the synthetic proofs already considered.
2. If one acute angle of a right triangle is double the other,
the hypotenuse is double the shorter side.
Given ZA = Z a, and ZC = Z2a, to prove that ^C is double BC.
Let M be the midpoint of AC. Then ^C is double BC if AM=BC.
Why ? Now if we draw MN II to CB, what can
be said of the relation of AN and NB ? Why ?
Then what may be said of A ANM and BNM?
Why ? Then what may be said of AM &nd BM ?
of Za and Zq? Therefore the proposition is
true if BM = BC. But BM= BC ii Z2a = Zr,
or if Z2 a = Za{ Zq, or it Za = Zq. But Za = Zq because we have
proved that AM = BM.
Now reverse this reasoning and write the proof in the usual synthetic
form.
METHODS OF PROOF 81
3. A median of a triangle is less than half the sum of the
two adjacent sides.
Given CM a median of the A ABC.
a
To prove that CM<i{BC \ CA).
Now CM<i{BC \CA),
if 2CM<BC + CA. "\ /J/
This suggests producing CM by its own length to P, \ /
and drawing AP. P
Then CP = 2 CM,
and 2 CM<BC + CA if CP<BC \CA.
But CP<^P + Cy1. Post. 3
.. CP<BC\ CA if BC = AP,
and BC = AP if A JlfBC is congraent to A MAP. § 67
But A MBC is congraent to A MAP, § 68
for MB = MA, Given
CM=MP, Hyp.
and ^JBJfC = Z^3fP. \ §60
.■.CP<BC{CA.
.'.CM<i{BC + CA).
4. The line which bisects two sides of a ti'iangle is parallel
to the third side. ^
Given AD equal to DB, and AE equal to EC.
To prove that DE is II to BC. Df
Suppose a Une drawn from C II to BA, and suppose DE
produced to meet it at G.
DE is II to BC if BCGD is a O.
BCGD is a O if CG = BD.
CG = BD if each is equal to AD.
Now BD = AD,
and CG = ^D if A CGE is congraent to A ADE.
But A CG^E is congraent to A ADE,
for EC = AE,
Z. CEG = Z AED,
and ZGCE = ZA.
82 BOOK I. PLANE GEOMETRY
5. Two isosceles triangles are congruent if a side and an
angle of the one are equal respectively to the corresponding
side and angle of the other.
The A are congruent if what three corresponding parts are equal ?
6. The bisector of an exterior angle of an isosceles tri
angle, formed by producing one of the equal
sides through the vertex, is parallel to the base.
AE is II to BC if what angles are equal ? These angles
are equal if Z CAD is twice what angle in the A ? ^ ^^
7. If one of the equal sides of an isosceles triangle is pro
duced through the vertex by its own length, the line joining
the end of the side produced to the nearer end of
the base is perpendicular to the base.
Z DBA is a rt. Z if it equals the sum of what A of A ABD ? ^
It equals this sum if Zp equals what angle and Z q equals
what other angle ?
8. If the equal sides of an isosceles triangle are produced
through the vertex so that the external segments are equal,
the extremities of these segments are equidistant from the
extremities of the base respectively.
9. If the line drawn from the vertex of a triangle to the
midpoint of the base is equal to half the base, the angle at
the vertex is a right angle.
10. If through any point in the bisector of an / ^^ ^
angle a line is drawn parallel to either side of
the angle, the triangle thus formed is isosceles, o
11. Through any point C in the line AJ3 an intersecting line
is drawn, and from any two points in this line equidistant from
C perpendiculars are drawn to AB ov AB produced. Prove that
these perpendiculars are equal.
12. The lines joining the midpoints of the sides
of a triangle divide the triangle into four congruent ^"^ ^"^
triangles.
METHODS OF PEOOF 83
157. The Indirect Method of Proof. The method of proof that
assumes the proposition false and then shows that this assump
tion is absurd is called the indirect method or the reductio ad
ahsurdum.
This method forms a kind of last resort in the proof of a proposition,
after the synthetic and analytic methods have failed.
EXERCISE 18
1. Given ABC and ABD, two triangles on the same base AB,
and on the same side of it, the vertex of each triangle being
outside the other triangle. Prove that ii AC equals
AD, then BC cannot equal BD.
Assume that BC = BD and show that the result is absurd,
since it would make D fall on C, which is contrary to the
given conditions.
2. On the sides of the angle XOY two equal segments OA
and OB are taken. On AB Si triangle APB is constructed with
AP greater than J5P. Prove that OP cannot y
bisect the angle XOY. ^ <r^'^^'^^
Assume that OP does bisect ZXOY. What is ^^^^J/"^
the result ? Is this result possible ? ^
3. From M, the midpoint of a line AB, MC is drawn oblique
to AB. Prove that CA cannot equal CB. c
Assume that CA does equal CB. What is the
result ? Is this result possible ?
4. If perpendiculars are drawn to the sides ^ m ^
of an acute angle from a point within the angle, they cannot
inclose a right angle or an acute angle.
Assume that they inclose a right angle and show that this leads to an
absurdity. Similarly for an acute angle.
5. One of the equal angles of an isosceles triangle is five
ninths of a right angle. Prove that the angle at the vertex
cannot be a right angle. 
Assume that it is a right angle. Is the result possible ?
84 BOOK I. PLANE GEOMETKY
158. General Suggestions for proving Theorems. The following
general suggestions will often be helpful :
1. Draw the figures as accurately as possible.
This is especially helpful at first. A proof is often rendered difficult
simply because the figure is carelessly drawn. If one line is to be laid off
equal to another, or if one angle is to be made equal to another, do this
by the help of the compasses or by measuring with a ruler.
2. Draw as general figures as possible.
If you wish to prove a proposition about a triangle, take a scalene tri
angle. If an equilateral triangle, for example, is taken, it may lead to
believing something true for every kind of a triangle, when, in fact, it
is true for only that particular kind.
3. After drawing the figure state very clearly exactly what
is given and exactly tvhat is to be proved.
Many of the difficulties of geometry come from failing to keep in mind
exactly what is given and exactly what is to be proved.
4. Then proceed synthetically with the proof if you see how
to begin. If you do not see how to begin, try the analytic method,
stating clearly that you could prove this if you could prove that,
and so on until you reach a known proposition.
5. If two lines are to be proved equal, try to prove them corre
sponding sides of congruent triangles, or sides of an isosceles
triangle, or opposite sides of a parallelogram, or segments between
parallels that cut equal segments from another transversal.
6. If two angles are to be proved equal, try to prove them
alternate interior or exteriorinterior angles of parallel lines, or
corresponding angles of congruent triangles, or base angles of
an isosceles triangle, or opposite angles of a parallelogram.
7. If one angle is to be j^^oved greater than another, it is prob
ably an exterior angle of a triangle, or an angle opposite the
greater side of a triangle.
8. If one line is to be proved greater than another, it is prob
ably opposite the greater angle of a triangle.
EXERCISES
85
EXERCISE 19
Prove the following propositions referring to equal lines :
1. If the sides AB and AD oi ?, quad
rilateral ABCD are equal, and if the di
agonal AC bisects the angle at A, then
EC is equal to DC.
A B
2. A line is drawn terminated by two parallel lines. Through
its midpoint any line is drawn terminated by the parallels.
Prove that the second line is bisected by the first.
3. In a parallelogram ABCD the line BQ
bisects AD, and DP bisects BC. Prove that Q,
BQ and DP trisect AC.
p X
4. On the base .45 of a triangle ABC any
point P is taken. The lines AP, PB, BC, and
CA are bisected by W, X, Y, and Z respec
tively. Prove that X F is equal to WZ. ^ w
5. In an isosceles triangle the medians drawn to the equal
sides are equal.
6. In the square ABCD, CD is bisected by Q, and P and R
are taken on AB so that AP equals BR. Prove that PQ
equals RQ. c
7. In this figure AC = BC, and ^P = BQ =
CR = CS. Prove that QR = PS.
8. From the vertex and the midpoints of the equal sides of
an isosceles triangle lines are drawn perpendicular to the base.
Prove that they divide the base into four equal parts.
9. In the quadrilateral ABCD it is known
that AB is parallel to DC, and that angle C
equals angle D. On CD two points are taken
such that CP=DQ. Prove that AP = BQ.
D Q
PC
c
86 BOOK I. PLANE GEOMETRY
EXERCISE 20
Prove the following propositions referring to equal angles :
1. In this figure it is given that AC = BC,
and that BQ and AR bisect the angles YBC
and CAX respectively. Prove that AAPB
is isosceles.
2. If through the vertices of an isosceles
triangle lines are drawn parallel to the oppo
site sides, they form an isosceles triangle.
3. If the vertical angles of two isosceles triangles coincide,
the bases either coincide or are parallel.
4. In which direction must the side of a
triangle be produced so as to intersect the
bisector of the opposite exterior angle ?
Consider the cases, ZA<ZC,ZA = ZC,ZA>ZC.
5. The bisectors of the equal angles of an isosceles triangle
form, together with the base, an isosceles triangle.
6. The bisectors of the base angles of an equilateral triangle
form an angle equal to the exterior angle at the
vertex of the triangle.
7. If the bisector of an exterior angle of a
triangle is parallel to the opposite side, the tri
angle is isosceles.
8. A line drawn parallel to the base of an isosceles triangle
makes equal angles with the sides or the sides produced.
9. A line drawn at right angles to AB, the base of an
isosceles triangle ABC, cuts ^C at P and BC produced at Q.
Prove that PCQ is an isosceles triangle. b d
10. In this figure, it AB = CD, and Z ^ = Z C, \ /
then BD is parallel to AC. \ J,
EXERCISES 87
EXERCISE 21
Prove the following propositions hy showing that two tri
angles are congruent :
1. A perpendicular to the bisector of an angle forms with
the sides an isosceles triangle.
2. If two lines bisect each other at right angles, any point in
either is equidistant from the extremities of the other,
3. From B a perpendicular is drawn to the bisector of the
angle A of the triangle ABC, meeting it at X, and meeting ^C
or AC produced at Y. Prove that BX = XY.
4. If through any point equally distant from two parallel
lines two lines are drawn cutting the parallels, they intercept
equal segments on these parallels. ^
5. If from the point where the bisector of an
angle of a triangle meets the opposite side,
parallels are drawn to the other two sides, and
terminated by the sides, these parallels are equal.
6. The diagonals of a square are perpendicular to each other
and bisect the angles of the square.
7. If from a vertex of a square there are drawn lineseg
ments to the midpoints of the two sides not adjacent to the
vertex, these linesegments are equal.
8. If either diagonal of a parallelogram bisects one of the
angles, the sides of the parallelogram are Q
all equal. / \^
9. On the sides of any triangle ABC equi / \^
lateral triangles BPC, CQA, ARB are con /^.^^/X
structed. Prove that AP = BQ = CR. ^\~~^— / ^ P
\ /'^
How can we prove that A ABP is congruent to \ /
ARBC? Also that A ^fiC is congruent to A ^^Q? V/
Does this prove the proposition ? \B " " ''
88 BOOK I. PLANE GEOMETRY
EXERCISE 22
Prove the following propositions relating to the sum of the
angles of a polygon :
1. An exterior angle of an acute triangle or of a right
triangle cannot be acute.
2. If the sum of two angles of a triangle equals the third
angle, the triangle is a right triangle.
3. If the line joining any vertex of a triangle to the mid
point of the opposite side divides the triangle into two isos
celes triangles, the original triangle is a right triangle.
4. If the vertical angles of two isosceles triangles are sup
plements one of the other, the base angles of the one are
complements of those of the other.
5. From the extremities of the base AB oi Si
triangle ABC perpendiculars to the other two sides
are drawn, meeting at P. Prove that the angle P is ^
the supplement of the angle C.
6. If two sides of a quadrilateral are parallel, and the other
two sides are equal but not parallel, the sums of the two pairs
of opposite angles are equal.
7. The bisectors of two consecutive angles of a parallelogram
are perpendicular to each other.
8. The exterior angles at B and C of any
triangle ABC are bisected by lines meeting
at P. Prove that the angle at P together
with half the angle A equals a right angle.
9. The opposite angles of the quadrilateral formed by
the bisectors of the interior angles of any quadrilateral are
supplemental.
10. Show that Ex. 9 is true, if the bisectors of the exterior
angles are taken.
EXERCISES 89
EXERCISE 23
Prove the following propositions referring to greater lines or
greater angles :
1. In the triangle ABC the angle A is bisected by a line
meeting BC at D. Prove that BA is greater than BD, and CA
greater than CD.
2. In the quadrilateral ABCD it is known that AD is the
longest side and BC the shortest side. Prove that the angle B
is greater than the angle D, and the angle C greater p
than the angle A. ^
3. A line is drawn from the vertex y4 of a square
ABCD so as to cut CD and to meet BC produced
in P. Prove that ^P is greater than DB.
4. If the angle between two adjacent sides of a parallelo
gram is increased, the length of the sides remaining unchanged,
the diagonal from the vertex of this angle is diminished.
c
5. Within a triangle ABC a point P is taken
such that CP = CB. Prove that AB is always
greater than A P.
^ A M B
6. In a quadrilateral ABCD it is known that AD equals BC
and that the angle C is less than the angle D. Prove that the
diagonal ^C is greater than the diagonal BD.
7. In the quadrilateral ABCD it is known that AD equals
BC and that the angle D is greater than the angle C. Prove
that the angle B is greater than the angle .4, c
8. In the triangle ABC the side AB is greater ^/
than A C. On A B and A C respectively BP is taken
equal to CQ. Prove that Z>Q is greater than CP. ^ ^
9. The sum of the distances of any point from
the three vertices of a triangle is greater than
half the sum of the sides.
90
BOOK I. PLANE GEOMETRY
EXERCISE 24
Prove the following miscellaneous exercises :
1. The line joining the midpoints of the nonparallel sides
of a trapezoid passes through the midpoints of jy
the two diagonals.
How is EF related to AB and BC ? Why ?
Since EF bisects BC and AB, how does it divide AC
aiidBB? Why?
2. The lines joining the midpoints of the
consecutive sides of any quadrilateral form
a parallelogram.
How are PQ and SR related to ^C ? ^ p b
3. If the diagonals of a trapezoid are equal, the trapezoid is
isosceles. c, d
Draw CE and BF ± to AB.
How isAABF related to A BCE ? Why ?
Then how is Z FAB related to Z CBA ?
Then how \sAABC related to A BAB? Why ? a E F B
4. If from the diagonal DB, of a square ABCD, BE is cut off
equal to 5C, and EF is drawn perpendicular to
BDj meeting DC 2X F, then DE is equal to jBi^ and
also to FC.
How many degrees in A EBF and BFE ? How is BE
related to J^F? Why?
Then how is rt. A BEF related to rt. A BCF ? Why ?
5. If the opposite sides of a hexagon are equal and parallel,
the diagonals that join opposite vertices meet in a point.
6. If perpendiculars are drawn from the four
vertices of a parallelogram to any line outside the
parallelogram, the sum of the perpendiculars from
one pair of opposite vertices equals the sum of
those from the other pair.
How are x ■\ y and to  2 related to A: ?
z
y A
k /y/
EXERCISES 91
EXERCISE 25
Examination Questions
1. The sum of the four sides of any quadrilateral is greater
than the sum of the diagonals.
2. The lines joining the midpoints of the sides of a square,
taken in order, form a square.
3. In a quadrilateral the angle between the bisectors of two
consecutive angles is one half the sum of the other two angles.
4. If the opposite sides of a hexagon are equal, does it follow
that they are parallel ? Give reasons for your answer.
5. In a triangle ABC the side BC is bisected at P and AB
is bisected at Q. AP is produced to R so that AP= PR, and
CQ is produced to S so that CQ = QS. Prove that S, B, and it
are in a straight line.
6. If the diagonals of a parallelogram are equal, all of the
angles of the parallelogram are equal.
7. In the triangle ABC, ZA = 60° and ZB>ZC. Which
is the longest and which is the shortest side of the triangle ?
Prove it.
8. How many sides has a polygon each of whose interior
angles is equal to 175° ?
9. Given the quadrilateral ABCD, with AB equal to AD,
and BC equal to CD. Prove that the diagonal A C bisects the
angle DCB and is perpendicular to the diagonal BD.
10. In how many ways can two congruent triangles be put
together to form a parallelogram ? Draw the diagrams.
11. The sides of a polygon of an odd number of sides are
produced to meet, thus forming a starshaped figure. What is
the sum of the angles at the points of the star ?
The propositions in Exercise 25 are taken from recent college entrance
examination papers.
92 BOOK I. PLANE GEOMETRY
EXERCISE 26
Review Questions
1. Define and illustrate rectilinear and curvilinear figures.
2. Upon what does the size of an angle depend ?
3. What is meant by the bisector of a magnitude ? Illus
trate when the magnitude is a line ; an angle.
4. Define perpendicular and state three facts relating to a
perpendicular to a line.
5. Name and define the parts of a triangle and such special
lines connected with a triangle as you have thus far studied.
6. Classify angles.
7. Classify triangles as to angles ; as to sides.
8. Define and illustrate complementary, supplementary, and
conjugate angles.
9. What are the two classes of assumptions in geometry ?
Give the list of each.
10. State all of the conditions of congruency of two triangles.
11. What is meant by the converse of a proposition ?
12. Are two triangles always congruent if three parts of the
one are respectively equal to three parts of the other ?
13. State three tests for determining whether one line is
parallel to another.
14. State the proposition relating to the sum of the angles of
a triangle, and state a proposition that can be proved by its use.
15. State a proposition relating to two unequal angles of a
triangle ; to two unequal sides of a triangle.
16. Must a triangle be equiangular if equilateral ? Must a
triangle be equilateral if equiangular ?
17. Classify polygons as to sides ; as to angles.
18. Define locus and give three illustrations.
BOOK II
THE CIRCLE
159. Circle. A closed curve lying in a plane, and such that
all of its points are equally distant from a fixed point in the
plane, is called a circle.
160. Circle as a Locus. It follows that the locus of a point in
a plane at a given distance from a fixed point is a circle.
161. Radius. A straight line from the center to the circle is
called a radius.
162. Equal Radii. It follows that all radii of the same circle
or of equal circles are equal, and that all circles of equal radii
are equal.
163. Diameter. A straight line through the center, termi
nated at each end by the circle, is called a diameter.
Since a diameter equals two radii, it follows that all diameters of the
same circle or of equal circles are equal.
164. Arc. Any portion of a circle is called an arc.
An arc that is half of a circle is called a semicircle.
An arc less than a semicircle is called a minor arc, and an arc greater
than a semicircle is called a major arc. The word arc taken alone is gen
erally understood to mean a minor arc.
165. Central Angle. If the vertex of an angle is at the center
of a circle and the sides are radii of the circle, the angle is
called a central angle.
An angle is said to intercept any arc cut off by its sides, and
the arc is said to subtend the angle.
94
BOOK II. PLANE GEOMETEY
Proposition I. Theorem
166. Ill the same circle or in equal ciixles equal cen
tral angles intercept equal arcs ; and of tivo unequal
central angles the greater intercepts the greater arc.
Given two equal circles with centers and 0\ with angles AOB
and A'O'B' equal, and with angle AOC greater than angle A'O'B^.
To prove that 1. arc AB = are A'B';
2. arc AC> arc A'B'.
Proof. 1. Place the circle with center O on the circle with
center 0' so that Z AOB shall coincide with its equal, AA'O'B'.
In the case of the same circle, swing one angle about until
it coincides with its equal angle.
Then A falls on A', and B on B'.
{Radii of equal circles are equal.)
.'. arc AB coincides with arc A'B'.
{Every point of each is equally distant from the center.)
Proof. 2. Since Z^OC is greater than Z^'0'5',
and ZAOB = Z.A 'O'B',
therefore . Z^OC is greater than Z. 4 (95.
Therefore OC lies outside ZAOB.
.'. 3iTG AC>3i,TGAB.
But ' arc ^5 = arc ^'^'.
.*. arc ^ C> arc A 'B', by Ax. 9. q.e.d.
Post. 5
§162
§159
Given
Given
Ax. 9
Ax. 11
CENTRAL ANGLES 95
Proposition II. Theorem
167. In the same circle or in equal circles equal arcs
subtend equal central angles ; and of two unequal arcs
the greater subtends the gi^eater central angle.
Given two equal circles with centers and O ', with arcs AB and
A^B^ equal, and with arc AC greater than arc 4 '5'.
To prove that 1. AAOB = Z. A'O'B';
2. ZAOOZ. A'O'B'.
Proof. 1. Using the figure of Prop. I, place the circle with
center O on the circle with center O' so that OA shall fall on
its equal O'A', and the arc AB on its equal A'B'. Post. 5
Then OB coincides with O'B'. Post. 1
.■.ZAOB = Z A'O'B'. §23
Proof. 2. Since arc AC> arc A'B', it is greater than arc AB,
the equal of arc A'B', and OB lies within the Z.AOC. Ax. 9
.■.ZA0OZ.A0B. Ax. 11
.\ZA0OZ A'O'B', by Ax. 9. q.e.d.
This proposition is the converse of Prop. I.
168. Law of Converse Theorems. Of four magnitudes, a, h, x, y,
if (1) a~>h when x > y,
(2) a = 5 when x = y,
and (3) a<h when x<y,
then the converses of these three statements are always true.
For when a > 6 it is impossible that x — y^ for then a would equal h
by (2) ; or that x < y, for then a would be less than 6 by (3). Hence x > y
when a > 6. In the same way, x = 2/ when a = h, and x <y when a <b.
169. Chord. A straight line that has its extremi ^■^^ss*^
ties on a circle is called a chord.
A chord is said to subtend the arcs that it cuts from a
circle. Unless the contrary is stated, the chord is taken
as subtending: the minor arc.
96 BOOK 11. PLANE GEOMETRY
Proposition III. Theorem
170. In the same circle or in equal circles, if two
arcs are equal, they are subtended hy equal chords ;
and if ttvo arcs are unequal, the greater is subtended
by the greater chord.
Given two equal circles with centers O and 0\ with arcs AB&nd
A'B^ equal, and with arc AF greater than arc A^B\
To prove that 1. chord AB = chord A' B' ;
2. chord AF> chord A' B'.
Proof. 1. Draw the radii OA, OB, OF, O'A', O'B'.
Since
OA = 0'A', and OB = O'B',
§162
and
ZAOB = ZA'0'B',
{In equal d) equal arcs siMend equal central A.)
§167
.'. A 0.45 is congruent to A O'A'B',
§68
and
chord .4 5 = chord .4 '5'.
§67
Proof.
2. In the A OAF and O'A'B',
OA = 0'A', and OF = O'B',
§162
but
Z A OF is greater than Z A 'O'B'.
§167
(In equal (D, of two unequal arcs the greater subtends the greater central Z. )
. • . chord AF> chord A 'B', by § 115. Q. e. d.
171. Corollary. In the same circle or in equal circles, the
greater of two unequal major arcs is subtended hy the less chord.
ARCS AND CHORDS 97
Proposition IV. Theorem
172. In the same circle or in equal circles, if two
chords are equal, they subtend equal arcs ; and if tivo
chords are unequal, the greater subtends the greater arc.
Given two equal circles with centers and 0\ with chords AB
and A'5' equal, and with chord AF greater than chord A'B\
To prove that 1. arc AB = arc A' B' ;
2. arcAF>arcA'B'.
Proof. 1. Draw the radii OA, OB, OF, O'A', O'B'.
Since OA = O'A', and OB = O'B', § 162
and chord AB = chord A 'B', Given
.'.AOABis congruent to A O'A'B', § 80
and ZAOB = ZA 'O'B'. § 67
.. arc AB = sltgA'B'. § 166
Proof. 2. In the A OAF and O'A'B',
OA = O'A ', and OF = O'B', § 162
but chord ^F> chord ^'5'. Given
.\ZAOF>ZA'0'B'. §116
. • . arc ^ jP > arc A 'B', by § 166. Q. e. d.
This proposition is the converse of Prop. III.
173. Corollary. In the same circle or in equal circles the
greater of two unequal chords subtends the less major arc.
98 BOOK 11. PLANE GEOMETRY
Proposition Y. Theorem
174. A line through the center of a circle perpendicular
to a chord bisects the chord and the arcs subtended by it.
Q
Given the line PQ through the center O of the circle AQBP^
perpendicular to the chord AB at M.
To prove that AM— BM, arc AQ = arc BQ, and arc AP =
arc BP.
Proof. Draw the radii OA and OB.
Then since OM = OM, Iden.
and OA = OB, § 162
.'. rt. A A MO is congruent to rt. A BMO. § 89
.\AM=BM, 3ind ZA0Q = ZQ0B. §67
Likewise Z POA = Z BOP. § 58
. • . arc AQ = arc BQ, and arc AP = arc BP, by § 166. Q. e. d.
175. Corollary 1. A diameter bisects the circle.
176. Corollary 2. A line through the center that bisects
a chord is perpendicular to the chord.
177. Corollary 3. The perpendicular bisector of a chord
passes through the center of the circle and bisects the arcs
subtended by the chord.
How many bisectors of the chord are possible ? How many ± bisec
tors ? Therefore with what line must this coincide (§ 174) ?
ARCS AND CHORDS 99
Proposition VI. Theorem
178. In the same circle or in equal circles equal chords
are equidistant from the center, and chords equidistant
from, the center are equal.
Given AB and CD, equal chords of the circle ACDB.
To prove that AB and CD are equidistant from the. center 0.
Proof. Draw OP ±toAB, and OQ _L to CD.
Draw the radii OA and OC.
OP bisects AB, and OQ bisects CD. § 174
Tlien since AP = CQ, Ax. 4
and OA = OC, § 162
.. rt. A OP A is congruent to rt. AOQC. § 89
.\OP = OQ. §67
.'. AB and CD are equidistant from 0, by § 88. q.e.d.
Given OP and OQ^ equal perpendiculars from the center O to the
chords AB and C2).
2^0 prove that AB = CD.
Proof. Since OA = OC, § 162
and OP = OQ, Given
.. rt. A OPA is congruent to rt. A OQC. § 89
.'.AP = CQ. §67
.. .4^ = C/>, by Ax. 3. Q.e.d.
100 BOOK II. PLANE GEOMETRY
Proposition VII. Theorem
179. In the same circle or in equal circles, if two
chords are unequal, they are unequally distant from
the center, and the greater chord is at the less distance.
Given a circle with center 0, two unequal chords AB and CD^
AB being the greater, and OP perpendicular to AB^ and OQ per
pendicular to CD.
To prove that OP <0Q.
Proof. Suppose AE drawn equal to CD, and OP, AtoAE.
Draw PR.
OP bisects AB, and OR bisects AE. § 174
{A line through the center of a circle ± to a chord bisects the chord.)
But AB> CD. Given
.. AB>AE, the equal of CD. Ax. 9
.. AP>AR. Ax. 6
.'.ZARP>ZRPA. §113
{If two sides of a A are unequal, the A opposite these sides are unequal,
and the Z opposite the greater side is the greater.)
.'. Z.PRO, the complement of ZARP, is less than Z.OPR,
the complement of Z RPA. § 59
.'.OP<OR. §114
But OR=OQ. §178
.. OP<OQ, by Ax. 9. Q.e.d.
AKCS AND CHORDS 101
Proposition VIII. Theorem
180. In the same circle or in equal circles, if tivo
chords are unequally distant from the center, they are
unequal, and the chord at the less distance is the greater.
Given a circle with center 0, two chords AB and CD unequally
distant from 0, and OP^ the perpendicular to AB^ less than OQ^
the perpendicular to CD.
To prove that AB>CD.
Proof. Suppose AE drawn equal to CD, and OR X to AE.
Now OP < OQ, Given
and OR = OQ. § 178
.. OP<OR. Ax. 9
Drawing PR, A PRO < A OPR. § 1 1 3
.•. Z ARP, the complement of Z PRO, is greater than Z RPA,
the complement of Z OPR. § 59
.\AP>AR. §114
But AP = \AB,2indiAR = \AE. §174
.'.AB>AE. Ax. 6
But CD = AE. Hyp.
.*. AB>CD, by Ax. 9. Q.e.d.
This proposition is the converse of Prop. VII.
181. Corollary. A diameter of a circle is greater than
any other chord.
102
BOOK II. PLANE GEOMETRY
182. Secant. A straight line that intersects a circle is called
a secant. In this figure AD is Si secant.
Since only two equal obliques can be drawn
to a line from an external point (§ 85), and
since the two equal angles which radii make
(§ 74) with any secant where it cuts the circle
cannot be right angles (§ 109), they must be
oblique ; and hence it follows that a secant can
intersect the circle in only two points.
183. Tangent. A straight line of unlimited length that has
one point, and only one, in common with a circle is called a
tangent to the circle.
In this case the circle is said to be tangent to the line. Thus in the
figure, BC is tangent to the circle, and the circle is tangent to BC.
The common point is called the point of contact or point of tangency.
By the tangent from an external point to a circle is meant the line
segment from the external point to the point of contact.
EXERCISE 27
1. A radius that bisects an arc bisects its sub
tending chord and is perpendicular to it.
2. On a circle the point P is equidistant from
two radii OA and OB. Prove that P bisects the
arc AB.
3. In this circle the chords AM and ]\fB are
equal. Prove that M bisects the arc AB and that
the radius OM bisects the chord AB.
4. On a circle are five points, A, B, C, D, E, so
placed that .45, BC, CD, DE are equal chords.
Prove that A C, BD, CE are equal chords, and
that AD and BE are also equal chords.
5. If two chords intersect and make equal angles
with the diameter through their point of intersec
tion, these chords are equal.
SECANTS AND TANGENTS 103
Proposition IX. Theorem
184. A line perpendicular to a radius at its extrem
ity is tangent to the circle.
A p
Given a circle, with XY perpendicular to the radius OP at P.
To prove that XY is tangent to the circle.
Proof. From draw any other line to XY, as OA.
Then OA>OP. §86
.'. the point A is outside the circle. § 160
Hence every point, except P, of the line AT is outside the
circle.
Therefore AT is tangent to the circle at P, by § 183. q.e.d.
185. Corollary 1. A tangent to a circle is perpendicular
to the radius drawn to the point of contact.
For OP is the shortest line from O to XY, and is therefore ± to XY
(§ 86) ; that is, XY is ± to OP.
186. Corollary 2. A perpendicular to a tangent at the
point of contact passes through the center of the circle.
For a radius is ± to a tangent at the point of contact, and therefore a
± erected at the point of contact coincides with this radius and passes
through the center of the circle.
187. Corollary 3. A perpendicular from the center of a
circle to a tangent passes through the point of contact.
What does § 86 say about this perpendicular ? '•■ .
104
BOOK II. PLANE GEOMETEY
188. Concentric Circles. Two circles that have the same center
are said to be concentric.
EXERCISE 28
1. The shortest chord that can be drawn through a given
point within a circle is that which is perpendicular
to the diameter through the poiiit.
Show that any other chord, CD, through P, is nearer
than is AB.
2. The diameter CD bisects the arc AB. Prove
that ZCi^^ =ABAC.
What kind of a triangle is A ^ JBC ?
3. Tangents at the extremities of a diameter
are parallel.
4. The arc AB is greater than the arc BC. OP
and OQ are perpendiculars from the center to AB
and BC respectively. Prove that Z QPO is greater
than Z OQP.
5. What is the locus of the center of a circle tangent to the
line XY dX the point P ? Prove it.
What two conditions must be shown to be fulfilled ?
6. What is the locus of the midpoints of a number of par
allel chords of a circle ? Prove it.
7. Three equal chords, AB, BC, CD, are placed /^^T^^^/?
end to end, and the radii OA, OB, OC, OD are 2)/
drawn. Prove that AAOC = AB0D.
8. All equal chords of a circle are tangent to a
concentric circle.
9. If a number of equal chords are drawn in
this circle, the figure gives the impression of a
second circle inside the first and concentric with
it. Explain the reason.
SECANTS AND TANGENTS 105
Proposition^ X. Theorem
189. Two parallel lines intercept equal arcs on a circle.
E~J!£^ F A
E
Fig. 1 Fig. 2 Fig. 3
Case 1. When the parallels are a tangent and a secant (Fig. 1).
Given AB^ a tangent at P, parallel to CD^ a secant.
To prove that arc CP = arc DP.
Proof. Suppose PP' drawn ± to AB Sit P.
Then PP' is a diameter of the circle. § 186
And PP' is also _L to CD. § 97
.. arc CP = arc Z)P. § 174
Case 2. When the parallels are both secants (Fig. 2).
Given AB and CD, parallel secants.
To prove that arc AC = arc BD.
Proof. Suppose EF II to CD and tangent to the circle at M.
Then arc ^3/= arc BM, and arc CM = arc DM. Case 1
.. arc ylC = arc 57). Ax. 2
Case 3. When the parallels are both tangents (Fig. 3).
Given AB^ a tangent at E, parallel to CD^ a tangent at F.
To prove that arc FGE = arc FHE.
Proof. Suppose a secant GH drawn II to AB.
Then arc GE = arc HE, and arc FG = arc FH. Case 1
.'. arc FGE = arc FHE, by Ax. 1. Q. e. d.
106 BOOK II. PLANE GEOMETRY
Proposition XI. Theorem
190. TJirough three points not in a straight line one
cii'cle, and only one, can he draion.
Given A, J?, C, three points not in a straight line.
To prove that one circle, and only one, can he drawn through
A, B, and C.
Proof. Draw AB and BC.
At the midpoints of AB and BC suppose Js erected.
These k will intersect at some point 0, since AB and BC are
neither parallel nor in the same straight line.
The point O is in the perpendicular bisector of AB, and is
therefore equidistant from A and B] the point is also in
the perpendicular bisector of BC, and is therefore equidistant
from B and C. § 150
Therefore is equidistant from A, B, and C.
Therefore a circle described about as a center, with a
radius OA, will pass through the three given points. § 160
The center of any circle that passes through the three points
must be in both of these perpendicular bisectors, and hence at
their intersection. As two straight lines can intersect in only
one point (§ 55), is the only point that can be the center of
a circle through the three given points. Q. e. d.
191 . Corollary. Tivo circles can intersect in only two points.
If two circles have three points in common, can it be shown that they
coincide and form one circle ?
SECANTS AND TANGENTS 107
Proposition XII. Theorem
192. Tlie tangents to a circle draivn from an external
point are equal, and tnake equal angles ivith the line
joining the point to the center.
A
Given PA and P5, tangents from P to the circle whose center is
0, and PO the line joining P to the center 0.
To prove that PA = PB, and ZAPO = Z OPB,
Proof. Draw OA and OB.
PA is ± to OA, and PB is ± to OB. § 185
{A tangent to a circle is J_ to the radius drawn to the point of contact.)
In the rt. A PAO and PBO,
PO = PO, Men.
and OA = OB. § 162
.. rt. A PA is congruent to rt. A PBO. § 89
.. PA = PB, and ZAPO = Z OPB, by § 67. Q. e. d.
193. Line of Centers. The line determined by the centers of
two circles is called the line of centers.
194. Tangent Circles. Two circles that are both tangent to the
same line at the same point are called tangent circles.
Circles are said to be tangent internally or externally, according as they
lie on the same side of the tangent Hne or on opposite sides. E.g. the
two circles shown in the figure on page 110 are tangent externally.
The point of contact with the line is called the point of contact or point
of tamjency of the circles.
108
BOOK II. PLANE GEOMETRY
EXERCISE 29
1. Show that the reasoning of § 190 will not hold for four
points, and hence that a circle cannot always
be drawn through four points.
2. Tangents to a circle 3it A, B, C, points
on the circle, meet in P and Q, as here shown.
Prove that AP \ QC = PQ.
3. If a quadrilateral has each side tangent to
a circle, the sum of one pair of opposite sides
equals the sum of the other pair.
In this figure, SP {■ QR = PQ + RS.
4. The hexagon here shown has each side
tangent to the circle. Prove that AB\ CD + EF
= BC ^ DE ^ FA.
5. In this figure CF is a diameter perpen
dicular to the parallel chords DB and EA, and
arc AB = 40° and arc 5C = 50°. How many de
grees are there in arcs CD, DE, EF, and FA ?
6. In this figure XYis tangent to the circle
at B, the chord CA is perpendicular to the j
diameter BD, and the arc CD = 150°. How
many degrees are there in arc AB?
7. If a quadrilateral has each side tangent to
a circle, the sum of the angles at the center
subtended by any two opposite sides is equal to a straight angle.
S. AP and CQ are parallel tangents meeting a
third tangent QP, as shown in the figure. be
ing the center, prove that the angle POQ is a
right angle.
Are A, 0, and C in the same straight line ? Draw OA
and OC, and find the relations of the zi at to those at P and Q.
LINE OF CENTERS 109
Proposition XIII. Theorem
195. If tivo circles intersect, the line of centers is the
perpendicular bisector of their common chord.
Given and 0', the centers of two intersecting circles, AB the
common chord, and 00' the line of centers.
To prove that 00' is A. to AB at its midpoint
Proof. Draw OA, OB, O'A, and O'B.
OA = OB, and O'A = O'B. § 162
.'. and O' are two points, each equidistant from A and B.
.*. 00' is the perpendicular bisector of AB, by § 151. q.e.d.
196. Common Tangents. A tangent to two circles is called a
common external tangent if it does not cut the linesegment
joining the centers, and a common internal tangent if it cuts it.
EXERCISE 30
Describe the relative position of two circles if the linesegment
joining the centers is related to the radii as stated in Exs. 15,
and illustrate each case hy a figure :
1. The linesegment greater than the sum of the radii.
2. The linesegment equal to the sum of the radii.
3. The linesegment less than the sum but greater than the
difference of the radii.
4. The linesegment equal to the difference of the radii.
5. The linesegment less than the difference of the radii.
110 BOOK 11. PLANE GEOMETRY
Proposition XIV. Theorem
197. If two circles are tangent to each other, the line
of centers passes through the point of contact.
A
B
Given two circles tangent at P.
To prove that P is in the line of centers.
Proof. Let AB be the common tangent at P. § 194
Then a J_ to ^4 B, drawn through the point P, passes through
the centers and 0\ § 186
{A l.to a tangent at the point of contact passes through the center
of the circle.)
Therefore the line determined by and 0', having two points
in common with this _L, must coincide with it. Post. 1
.*. P is in the line of centers. q.e.d.
EXERCISE 31
Describe the relative position of two circles having tangents
as stated in Exs. 1~5, and illustrate each case by a figure :
1. Two common external and two common internal tangents.
2. Two common external tangents and one common internal
tangent.
3. Two common external tangents and no common internal
tangent.
4. One common external and no common internal tangent.
5. No common tangent.
TANGENTS 111
6. The line which passes through the midpoints of two
parallel chords passes through the center of the circle.
7. If two circles are tangent externally, the tangents to them
from any point of the common internal tangent are equal.
8. If two circles tangent externally are tangent to a line
AB at A and B, their common internal tangent bisects AB.
9. The line drawn from the center of a circle to the point
of intersection of two tangents is the perpendicular bisector of
the chord joining the points of contact.
10. The diameters of two circles are respectively 2.74 in. and
3.48 in. Find the distance between the centers of the circles
if they are tangent externally. Find the distance between the
centers of the circles if they are tangent internally.
11. Three circles of diameters 4.8 in., 3.6 in., and 4.2 in. are
externally tangent, each to the other two. Find the perimeter
of the triangle formed by joining the centers.
12. A circle of center and radius r' rolls around a fixed
circle of radius r. What is the locus of O? Prove it.
13. The line drawn from the midpoint of a chord to the
midpoint of its subtended arc is perpendicular to the chord.
14. If two circles tangent externally at P are tangent to a
line AB at A and B, the angle BPA is a right angle.
15. Three circles are tangent externally at the points A, B,
and C, and the chords AB and AC are produced to cut the
circle BC Sit D and E. Prove that DE is a diameter.
16. If two radii of a circle, at right angles to each other,
when produced are cut by a tangent to the circle at A and B,
the other tangents from A and B are parallel to each other.
17. If two common external tangents or two common inter
nal tangents are drawn to two circles, the linesegments inter
cepted between the points of contact are equal.
112 BOOK 11. PLANE GEOMETRY
198. Measure. The number of times a quantity of any kind
contains a known unit of the same kind, expressed in terms of
that known imit, is called the measure of the quantity.
Thus we measure the length of a schoolroom by finding the number of
times it contains a known unit called the foot. We measure the area of
the floor by finding the number of times it contains a known unit called
the square foot. You measure your weight by finding the number of
times it contains a known unit called the pound. Thus the measure of
the length of a room may be 30 ft., the measure of the area of the floor
may be 600 sq. ft., and so on.
The abstract number found in measuring a quantity is called
its numerical tneasui^e, or usually simply its measure.
199. Ratio. The quotient of the numerical measures of two
quantities, expressed in terms of a common unit, is called the
ratio of the quantities.
Thus, if a room is 20 ft. by 35 ft., the ratio of the width to the length
is 20 ft, = 35 ft., or , which reduces to ^. Here the common unit is 1 ft.
The ratio of a to 6 is written , or a : 6, as in arithmetic and algebra.
b
Thus the ratio of 20° to 30° is f§, or , or 2 : 3.
200. Commensurable Magnitudes. Two quantities of the same
kind that can both be expressed in integers in terms of a com
mon unit are said to be commensurable magnitudes.
Thus 20 ft. and 35 ft. are expressed in integers (20 and 35) in terms
of a common unit (1 ft.); similarly 2 ft. and 3 ft., the integers being
4 and 7, and the common unit being \ ft.
The common unit used in measuring two or more commensurable
magnitudes is called their common measure. Each of the magnitudes is
called a multiple of this common measure.
201. Incommensurable Magnitudes. Two quantities of the
same kind that cannot both be expressed in integers in terms
of a common unit are said to be incommensurable magnitudes.
Thus, if a = V2 and 6 = 3, there js no number that is contained an
integral number of times in both V2 and 3. Hence a and h are, in this
case, incommensurable magnitudes.
measureme:n^t 113
202. Incommensurable Ratio. The ratio of two incommensur
able magnitudes is called an incommensurable ratio.
Although the exact value of such a ratio cannot be expressed
by an integer, a common fraction, or a decimal fraction of a
limited number of places, it may be expressed approximately.
Thus suppose  = V2.
Now V2 =1.41421356 •••, which is greater than 1.414213
but less than 1.414214. Then if a millionth part of b is taken
as the unit of measure, the value oi a:b lies between 1.414213
and 1.414214, and therefore differs from either by less than
0.000001.
By carrying the decimal further an approximate value may
be found that will differ from the ratio by less than a billionth,
a trilllonth, or any other assigned value.
That is, for i^ractical purposes all ratios are commensurable.
For example, if  > — but < , then the error in taking either of
1)71 n
these values for  is less than  , the difference of these ratios. But by
b 1 ""
increasing n indefinitely,  can be decreased indefinitely, and a value of
the ratio can be found within any required degree of accuracy.
EXERCISE 32
Find a common measure of :
1. 32 in., 24 in. 3. 5^ in., 3^ in. 5. 6^ da., 2 da.
2. 48 ft, 18 ft. 4. 2 lb., 1^ lb. 6. 14.4 in., 1.2 in.
Find the greatest common measure of:
7. 64 yd., 24 yd. 9. 7.5 in., 1.25 in. 11. 2 ft, 0.25 ft
8. 51 ft., 17 ft 10. 31 in., 0.33J in. 12. 75°, 7° 30'.
13. lia:b = V3, find an approximate value of this ratio that
shall differ from the true value by less than 0.001.
114
BOOK II. PLANE GEOMETKY
203. Constant and Variable. A quantity regarded as having
a fixed value throughout a given discussion is called a constant,
but a quantity regarded as having different successive values
is called a variable.
204. Limit. When a variable approaches a constant in such
a way that the difference between the two may become and
remain less than any assigned positive quantity, however
small, the constant is called the Ihnit of the variable.
Variables can sometimes reach their limits and sometimes not. E.g. a
chord may increase in length up to a certain limit, the diameter, and
it can reach this limit and still be a chord ; it may decrease, approaching
the limit 0, but it cannot reach this limit and still be a chord.
205. Inscribed and Circumscribed Polygons. If the sides of a
polygon are all chords of a circle, the polygon is said to be
inscribed in the circle ; if the sides are all tangents to a circle,
the polygon is said to be circum
scribed about the circle.
The circle is said to be circum
scribed about the inscribed polygon,
and to be inscribed in the circum
scribed polygon.
Inscribed
Polygon
Circumscribed
Polygon
206. Circle as a Limit. If we inscribe a square in a circle,
and then inscribe an octagon by taking the midpoints of the
four equal arcs for the new vertices, the octa
gon is greater than the square but smaller than
the area inclosed by the circle, and the perim
eter of the octagon is greater than the perim
eter of the square (§ 112).
By continually doubling the number of sides
in this way it appears that the area inclosed by the circle is the
limit of the area of the polygon, and the circle is the limit of
its perimeter, as the number of sides is indefinitely increased.
Hence we have limiting forms as well as limiting values, the form of
the circle being the limit approached by the form of the inscribed polygon.
LIMITS
115
\L
■\M
207. Principle of Limits. If, ivhile approaching their respec
tive limits, two variables are always equal, their limits are equal.
Let AX and BY increase in
length in such a way that they
always remain equal, and let
their respective limits be ylL
and BM.
To prove that AL = BM.
Suppose these limits are not equal, but that AZ = BM.
Then since X may reach a point between Z and L we may
have AX> AZ, and therefore greater than its supposed equal,
BM; but 5 F cannot be greater than BM. Therefore we should
have AX> BY, which is contrary to what is given.
Hence BM cannot be greater than AL, and similarly AL
cannot be greater than BM. .'. AL = BM. q.e.d.
208. Area of Circle. The area inclosed by a circle is called
the area of the circle.
It is evident that a diameter bisects the area of a circle.
209. Segment. A portion of a plane bounded by an arc of a
circle and its chord is called a segment
of the circle.
If the chord is a diameter, the segment is
called a semicircle, this word being commonly
used to mean not only half of the circle but also
the area inclosed by a semicircle and a diameter.
210. Sector. A portion of a plane ^m,c.b
bounded by two radii and the arc of the circle intercepted by
the radii is called a sector.
If the arc is a quarter of the circle, the sector is called a quadrant.
211. Inscribed Angle. An angle whose vertex is on a circle,
and whose sides are chords, is called an inscribed angle.
An angle is said to be inscribed in a segment if its vertex is on the arc
of the segment and its sides pass through the ends of the arc.
116 BOOK II. PLANE GEOMETRY
Proposition XV. Theorem
212. In the same circle or in equal circles tivo central
angles have the same ratio as their intercepted arcs.
Fig. 1 Fig. 2 Fig. 3
Given two equal circles with centers and 0\ AOB and A'O'B'
being central angles, and AB and A^B' the intercepted arcs.
To prove that , , ^ — =
Case 1. When the ares are commensurahle (Figs. 1 and 2).
Proof. Let the arc m be a common measure of A^B' and AB.
Apply the arc m as a measure to the arcs A'B^ and ^5 as
many times as they will contain it.
Suppose m is contained a times in A'B', and b times in AB.
Then
arc A *B' a
.SiVcAB b
At the several points of division on AB and A 'B' draw radii.
These radii will divide Z.AOB into b j)arts, and Z.A'0'B'
into a parts, equal each to each. § 167
ZA'O'B' a
ZAOB ~1j'
ZA'O'B' slygA'B' , , „
.. ■ ^ ^^ = — J by Ax. 8. Q.E.D.
ZAOB SiYcAB ^
Case 2 may be omitted at the discretion of the teacher if the incom
mensurable cases are not to be taken in the course.
MEASURE OF ANGLES 117
Case 2. When the arcs are incommensurable (Figs. 2 and 3).
Proof. Divide AB into a number of equal parts, and apply
one of these parts to A 'B' as many times as A 'B' will contain it.
Since AB and A'B' are incommensurable, a certain number
of these parts will extend from A' to some point, as P, leaving
a remainder PB' less than one of these parts. Draw O'P.
By construction AB and A'P are commensurable.
ZA'O'P arc^'P
* ' ZAOB ^vgAB
Case 1
By increasing the number of equal parts into which AB is
divided we can diminish the length of each, and therefore can
make PB' less than any assigned positive value, however small.
Hence PB' approaches zero as a limit as the number of parts
of AB is indefinitely increased, and at the same time the
corresponding angle PO'B' approaches zero as a limit. § 204
Therefore the arc ^'P approaches the arc A'B' as a limit, and
the A A' O'P approaches the Z A'O'B' as a limit.
. , , arc^'P , arcyl'5' .. .^
.*. the variable — approaches — — as a Innit,
arc^^ arc .45
• VI AA'O'P , Z. A'O'B' .. ..
and the variable , , ,,„ approaches . . __ as a limit.
Z.AOB ^^ AAOB
ZA'O'P . , , ^ arc^'P
But . , ^,, IS always equal to — >
Z.AOB ./ ^ siTcAB
as A'P varies in value and approaches A'B' as a limit. Case 1
Z A'O'B' SiYcA'B' , . OAT .^T^
.'.^ = — jby§207. Q.E.D.
ZAOB SiTcAB ^
213. Numerical Measure. We therefore see that the numerical
measure of a central angle (in degrees, for example) equals the
numerical measure of the intercepted arc. This is commonly
expressed by saying that a central angle is measured by the
intercepted arc.
118 BOOK 11. PLANE GEOMETRY
Proposition XVI. Theorem
214. A7i inscribed angle is measured hy half the in
tercepted arc.
B B B
Given a circle with center O and the inscribed angle B^ inter
cepting the arc AC.
To prove that Z. B is measured hy half the arc A C.
Case 1. When is on one side, as AB (Fig. 1).
Proof. Draw OC.
Then '.'OC = OB, §162
.\AB = AC. §74
But Z.B\AC = Z. A OC. § 111
.•.2AB = AA0C. Ax. 9
.'.Z.B = :^ Z.AOC. Ax. 4
But AAOC is measured by arc A C. § 213
.*. 1 Z. AOC is measured by ^ arc AC. Ax. 4
.'. Z.B is measured by ^ arc AC. Ax. 9
Case 2. When lies within the angle B (Fig. 2).
Proof. Draw the diameter BD,
Then Z ABD is measured by \ arc AD,
and Z DBC is measured by ^ arc DC. Case 1
.. Z ^57) + Z Z)5C is measured by \ (arc ^i) + arc i)C),
or Z ^^C is measured by \ arc AC.
MEASURE OF ANGLES
119
Case 3. W?ien lies outside the angle B (Fig. 3).
Proof. Draw the diameter BD.
Then Z DBC is measured by \ arc 7)C,
and Z DBA is measured by \ arc DA. Case 1
.'. /.DBC — Z. DBA is measured by \ (arc DC — arc /).!),
or
Z.ABC is measured by \ arc .4C.
Q.E.D.
Fig. 4
Fig. 6
215. Corollary 1. An angle inscribed in a semicircle is a
right angle.
For it is half of a central straight angle, as,in Fig, 4.
216. Corollary 2. An angle inscribed in a segment greater
than a semicircle is an acute angle, and an angle inscribed in
a segment less than a semicircle is an obtuse angle.
See A A and B in Fig, 5.
217. Corollary 3. Angles inscribed in the same segment
or in equal segments are equal.
Why is this ? (Fig. 6.)
218. Corollary A:. If a quadrilateral is inscribed in a
circle, the opposite angles are supplementary ; and, conversely,
if two opposite angles of a quadrilateral are supplementary,
the quadrilateral can be inscribed in a circle.
For the second part, can a circle be passed through A^ /f^f/^X
jB, C (§ 190) ? If it does not pass through D also, can you f I d>~^A
show that Z D would be greater than or less than some other jj^^^^^"^
angle (§111) that is supplementary to Z S ?
120 BOOK II. PLAXE GEOMETRY
EXERCISE 33
1. A parallelogram inscribed in a circle is a rectangle.
2. A trapezoid inscribed in a circle is isosceles.
3. The shorter segment of the diameter through a given
point within a circle is the shortest line that can
be drawn from that point to the circle. f o p^
Let P be the given point. Prove PA shorter than any
other line PX from P to the circle.
4. The longer segment of the diameter through a given point
within a circle is the longest line that can be drawn from that
point to the circle.
5. The diameter of the circle inscribed in
a right triangle is equal to the difference
between the hypotenuse and the sum of
the other two sides.
6. A line from a given point outside a circle passing through
the center contains the shortest linesegment that can be drawn
from that point to the circle.
Let P be the point, the center, A the point
where PO cuts the circle, and C any other point on ^'
the circle. How does PC\CO compare with P02
. 7. A line from a given point outside a circle passing through
the center contains the longest linesegment (to the concave
arc) that can be drawn from that point to the circle.
8. Through one of the points of intersection of two circles
a diameter of each circle is drawn. Prove that
the line joining the ends of the diameters passes
through the other point of intersection.
9. If two circles intersect and a line is drawn
through each point of intersection terminated by
the circles, the chords joining the corresponding
ends of these lines are parallel.
MEASURE OF ANGLES 121
Proposition XVIT. Theorem
219. A71 angle formed hy two chords intersecting
ivithin the circle is measured hy half the sum of the
intercepted arcs.
Given the angle AOB formed by the chords AC and BD.
To prove that Z A OB is measured hy ^ (ar<? AB f arc CD^.
Proof. ' Draw A D.
Then AAOB = AA+Z.D. §111
{An exterJ,or Z 0/ a A is equal to the sum of the two opposite interior A.)
But Z A is measured by ^ arc CD, § 214
{An inscribed Z is measured by half the intercepted arc.)
and Z Z) is measured by ^ arc AB. § 214
.. Z. AOB is measured by J (arc A B + arc CD), by Ax. 1. Q. e. d.
Discussion. If is at the center of the circle, to what previous prop
osition does this proposition reduce ?
If is on the circle, as at U, to what previous proposition does this
proposition reduce ?
Suppose the point remains as in the figure, and the chord AC
swings ahout as a pivot until it coincides with the chord BD. What
can then be said of the measure of A AOB and COD ? What can be said
as to the measure oi ABOC and DO A ?
It is also possible to prove the proposition by drawing a chord AE
parallel to BD, and showing that ZAOB = ZA, since they are alternate
interior angles formed by a transversal cutting two parallels. Now Z A
is measured by I arc CE. But arc CE = arc CD + arc DE^ or arc CD +
arc ^B, since arc AB = arc DE (§ 189). Therefore ZAOB, which equals
ZA, is measured by i (arc AB \ arc CD).
122 BOOK II. PLAKE GEOMETRY
Propositiox XVIII. Theorem
220. An angle formed hy a tangent and a chord drawn
from the point of contact is measured hy half the in
tercepted arc.
Given the chord PQ and the tangent XY through P.
To prove that A QPX is measured hy half the are QSP.
Proof. Suppose the chord QPi, drawn from the point Q par
allel to the tangent XY.
Then arc PA' = arc QSP. * § 189
{Two parallel lines intercept equal arcs on a circle.)
Also Z QPX = Z PQR. § 100
{If two parallel lines are cut by a transversal, the alternateinterior
angles are equal.)
But Z PQR is measured by J arc PR. § 214
{An inscribed Z is measured by half the intercepted arc.)
Substitute Z QPX for its equal, the Z PQR,
and substitute arc QSP for its equal, the arc PR.
Then Z QPX is measured by J arc QSP, by Ax. 9. q.e.d.
Discussion. By half of what arc is Z FPQ, the supplement of Z QPX,
measured ?
If PQ should be drawn so as to be perpendicular to XY, by what
would A YPQ and QPX be measured ?
Suppose PQ swings about the point P as a pivot until it coincides
with XY, by what will Z YPQ be measured ? By what will Z QPX be
measured, and what will it equal ?
MEASURE OF ANGLES
123
Proposition XIX. Theorem
221. An angle formed hy two secants, a secant and
a tangent, or tivo tangents, drawn to a circle from an
external point, is measured hy half the difference of
the intercepted arcs.
Fig. 1
Fig. 3
Given two secants PBA and PCD^ from the external point P.
To prove that Z P is measured hy ^ (arc DA — arc BC),
Proof. Suppose the chord BX drawn II to PCD (Fig. 1).
Then
Furthermore
§189
Ax. 9
§102
§214
arc BC = arc DX.
arc XA = arc DA — arc DX.
.'. arc XA — arc DA — arc BC.
Also ZP = ZXBA.
But Z XBA is measured by ^ arc XA.
Substitute AP for its equal, the Z.XBA,
and substitute arc DA — arc BC for its equal, the arc XA.
Then Z P is measured by J (arc DA — arc BC), hy Ax. 9. Q. e. d.
If the secant PBA Y swings around to tangency, it becomes
the tangent PB and Fig. 1 becomes Fig. 2. If the secant PCD
also swings around to tangency, it becomes the tangent PC and
Fig. 2 becomes Fig. 3. The proof of the theorem for each of
these cases is left for the student.
124
BOOK 11. PLANE GEOMETRY
EXERCISE 34
1. If two circles touch each other and two lines are drawn
through the point of contact terminated by the circles, the chords
joining the ends of these lines are parallel.
This could be proved if it could be shown that
Z A equals what angle ? To what two angles can
these angles be proved equal by § 220 ? Are those
angles equal ?
2. If one side of a right triangle is the diameter of a circle,
the tangent at the point where the circle cuts the hypotenuse
bisects the other side.
If OE is II to J.C, then because BO = OA, what is the
relation of BE to EC ? The proposition therefore reduces
to proving that OE is parallel to what line ? This can be
proved if Z BOE can be shown equal to what angle ?
3. If from the extremities of a diameter AD
two chords, A C and DB, are drawn intersecting
at P so as to make Z APB = 45°, then Z BOC ^
is a right angle.
4. The radius of the circle inscribed in an equilateral tri
angle is equal to one third the altitude of the ^
triangle.
To prove this we must show that AF equals what
line? It looks as if AF might equal EF, and EF
equal OF. Is there any way of proving A OFE equi
lateral ? of proving A AEF isosceles ?
5. If two lines are drawn through any point in a diagonal
of a square parallel to the sides, the points where these lines
meet the sides lie on the circle whose center
is the point of intersection of the diagonals.
OY = OZ if what two A are congruent? Why are
these A congruent ? OY = OX if what two A are con
gruent ? OX = 0W if what two A are congruent ? a
PRINCIPLE OF CONTINUITY
125
222. Positive and Negative Quantities. In
geometry, as in algebra, quantities may be
distinguished as positive and as negative.
Thus as we consider temperature above zero posi
tive and temperature below zero negative, so in this
figure, if OB is considered positive, then OD may be
considered negative. Similarly, if OA is considered
positive, then OC may be considered negative.
Likewise with respect to angles and arcs, if the
rotating line OA moves in the direction of AB,
counterclockwise, the angle and arc generated are
considered positive. If it rotates in the direction
AB', like the hands of a clock, the angle and arc
generated are considered negative.
223. Principle of Continuity. By considering the distinction
between positive and negative magnitudes, a theorem may
often be so stated as to include several particular theorems.
For example. The angle included between two lines that cut or
are tangent to a circle is measured by half the sum of the
intercepted arcs.
In particular : 1. If the lines intersect at the center, half the sum of
the arcs will then become simply one of the arcs, and the proposition
reduces to that of § 213.
2. If the lines are two general chords, we have the case of § 219.
3. If the point of in
tersection P moves to the
circle, we have the case ( \/ \ f 2\ \ i /P\
of § 214, one arc becom
ing zero.
4. If P moves outside the circle, then the smaller arc passes through
zero and becomes negative, so that the sum of the arcs becomes their
arithmetical difference (§ 221).
We may continue the discussion so as to include all the cases of
the propositions proved from § 213 to § 221.
When the reasoning employed to prove a theorem is con
tinued as just illustrated, so as to include several theorems,
we are said to reason by the Principle of Continuity.
126 BOOK II. PLANE GEOMETRY
224. Problems of Construction. At the beginning of the study
of geometry some directions were given for simple construc
tions, so that figures might be drawn with accuracy. It was
not proved at that time that these constructions were correct,
because no theorems had been studied on which proofs could
be based. It is now purposed to review these constructions, to
prove that they are correct, and to apply the methods employed
to the solution of more difficult problems.
225. Nature of a Solution. A solution of a problem has one
requirement that a proof of a theorem does not have.
In a theorem we have three general steps : (1) Given, (2) To
prove, (3) Proof. In a problem we have four steps : (1) Given,
(2) Required (to do some definite thing), (3) Construction (show
ing how to do it), (4) Proof (that the construction is correct).
We prove a theorem, but we solve a problem, and then prove
that our solution is a correct one.
In the figures of this text given lines are shown as full, black lines ;
construction lines and lines required are shown as dotted lines.
226. Discussion of a Problem. Besides the four necessary
general steps in treating a problem, there is a desirable step
to be taken in many cases. This is the discussion of the prob
lem, in which is considered whether there is more than one
solution, and other similar questions.
For example, suppose the problem is this : Required from a given point
to draw a tangent to a circle.
After the problem has been solved we may discuss it thus : In general,
if the given point is outside the circle, two tangents may be drawn,
and these tangents are equal (§ 192) ; if the given point is on the circle
only one tangent can be drawn, since only one perpendicular can be
drawn to a radius at its extremity (§ 184) ; if the given point is within
the circle, evidently no tangent can be drawn.
In the discussion the Principle of Continuity often enters, the figure
being studied for various positions of some given point or line, as was
done in the discussions on pages 121 and 122.
PKOBLEMS OF CONSTRUCTION 127
Proposition XX. Problem
227. To let fall a perpendicular iq^on a given line
from a given external point.
P\
Yy
Given the line AB and the external point P.
Required from P to let fall a J_ upon AB.
Construction. With P as a center, and a radius sufficiently
great, describe an arc cutting ^1^ at J^ and Y. Post. 4
With X and Y as centers, and a convenient radius, describe
two arcs intersecting at C. Post. 4
Draw PC. Post. 1
Produce PC to intersect AB at M. Post. 2
Then PM is the line required. Q. e. f.
Proof. Since P and C are by construction two points each
equidistant from X and F, they determine the perpendicular
to Jl F at its midpoint. § 151
{Two points each equidistant from the extremities of a line determine
the ± bisector of the line.) Q. E. D.
Discussion. The following are interesting considerations :
That PC produced will really intersect AB, as stated in the construc
tion, is shown in the proof.
A convenient radius to take for the two intersecting arcs is XY.
If C falls on P, take C at the other intersection of the arcs below AB^
as is seen in the figure of Ex. 2, p, 9.
To obtain a radius for the first circle, draw any line from P that will
cut AB, and use that.
128 BOOK II. PLANE GEOMETRY
Proposition XXI. Problem
228. At a given point in a given line, to erect a per
pendicular to the line,
•>.a — . \y
I
! \
Oy
/
^'
^ \X P Yl " X^^.
Fig. 1 Fig. 2
Given the point P in the line AB.
Required to erect a A. to AB at P.
Case 1. When the point P is not at the end of AB (Fig. 1).
Construction. Take PX = PY. Post. 4
With X and Y as centers, and a convenient radius, describe
arcs intersecting at C. Post. 4
Draw CP. Post. 1
Then CP is ±to AB. q.e.f.
Proof. P and C, two points each equidistant from X and Y,
determine the ± bisector of ZF, by § 151. q.e.d.
Case 2. When the point P is at the end of AB (Fig. 2).
Construction. Suppose P to coincide with B.
Take any point outside of AB, and with as a center and
OB as a radius describe a circle intersecting AB a,t X.
From X draw the diameter XY, and draw BY. Post. 1
Then i^F is ± to AB. Q.e.f.
Proof. Z 5 is a right angle. § 215
.. BY is ± to AB, by § 27. Q.e.d.
Discussion. If the circle described with as a center is tangent to
AB ?it B, then OB is the required perpendicular (§185).
PROBLEMS OF CONSTRUCTION
Proposition XXII. Problem
229. To bisect a given line.
I
129
■B
M
Given the line AB.
Required to bisect AB.
Construction. With A and B as centers and AB as a radius
describe arcs intersecting at X and Y, and draw XY. Post. 4
Then XY bisects AB. q.e.f.
Proof. XY bisects AB, by § 151. q.e.d.
Proposition XXIII. Problem
230. To bisect a given arc.
\g
Given the arc AB.
Required to bisect AB.
Construction. Draw the chord AB. Post. 1
Draw CM, the perpendicular bisector of the chord AB. § 229
Proof.
Then CM bisects the arc AB.
CM bisects the arc AB, by § 177.
Q.E.F.
Q.E.D.
130 BOOK 11. PLANE GEOMETRY
Proposition XXIV. Problem
231. To bisect a given angle.
o
Given the angle AOB.
Required to bisect A AOB.
Construction. With O as a center and any radius describe an
arc cutting OA at X and OB at Y. Post. 4
With X and Y as centers and .Y F as a radius describe arcs
intersecting at P. Post. 4
Draw OP. Post. 1
Then OP bisects Z.AOB. q.e.f.
Proof. Draw PX and PY.
Then prove that the A OXP and YP are congruent. § 80
Then AA0P = Z.P0B,hj4 67. Q. e. d.
EXERCISE 35
1. To construct an angle of 45°; of 135°.
2. To construct an angle of 22° 30'; of 157° 30'.
3. To construct an equilateral triangle, having given one
side, and thus to construct an angle of 60°.
4. To construct an angle of 30° ; and thus to trisect a right
angle.
5. To construct an angle of 15°; of 7° 30'; of 195°; of 345°.
6. To construct a triangle having two of its angles equal
to 75°. Is the triangle definitely determined ?
PROBLEMS OF CONSTRUCTIOISr 131
Proposition XXV. Problem
232. From, a given point in a given line, to draw a
line making an angle equal to a given angle.
p
Ic
\m *
Given the angle AOB and the point P in the line PQ.
Required from P to draw a line making with the line PQ
an angle equal to A A OB.
Construction. With O as a center and any radius describe an
arc cutting OA at C and OB at D. Post. 4
With P as a center and the same radius describe an arc MX^
cutting PQ at M, Post. 4
With M as a center and a line joining C and D as a radius
describe an arc cutting the arc MX at N. Post. 4
Draw PN. Post. 1
Then Z QPN = Z A OB. Q. e. f.
Proof. Draw CD and MN.
Then prove that the A PMN and OCD are congruent. § 80
Then Z QPN =Z.A OB, by § 67. Q. e. d.
233. Corollary. Through a given external point, to draiv
a line parallel to a given line.
X
Let AB be the given hne and P the given external „/'
point. C f^D
Draw any Une XPY through P, cutting AB as in /q
the figure. /
Draw CD through P, making Z.p = Zq. /
The Hne CD will be the line required.
132 BOOK II. PLANE GEOMETRY
Proposition XXVI. Problem
234. To divide a given line iiito a given number of
equal parts.
Ai^ 1 ; iB
/^ 7 7 7
' ^^. /
Given the line AB.
Required to divide AB into a given number of equal parts.
Construction. From A draw the line AO, making any con
venient angle with AB. Post. 1
Take any convenient length, and by describing arcs apply
it to ^ as many times as is indicated by the number of parts
into which ^^ is to be divided. Post. 4
From C, the last point thus found on A 0, draw CB. Post. 1
From the division points on ^ draw parallels to CB. § 233
These lines divide AB into equal parts. q.e.f.
Proof. These lines divide AB into equal parts, by § 134. q. e. d.
EXERCISE 36
1. To divide a given line into four equal parts.
2. To construct an equilateral triangle, given the perimeter.
3. Through a given point, to draw a line which shall make
equal angles with the two sides of a given angle.
4. Through a given point, to draw two lines so that they
shall form with two intersecting lines two isosceles triangles.
5. To construct a triangle having its three angles respec
tively equal to the three angles of a given triangle.
PKOBLEMS OF CONSTRUCTION 133
Proposition XXVII. Problem
235. To construct a triangle lohen two sides and the
included angle are given.
4. :i— X
;c B
Given h and c two sides of a triangle, and O the included angle.
Required to construct the triangle.
Construction. On any line as AX, by describing an arc, mark
o^ AB equal to c. Post. 4
At A construct A BAD equal to Z 0. § 232
On AD, by describing an arc, mark off ^ C equal to h. Post. 4
Draw BC. Post. 1
Then A ABC is the A required. q.e.f.
Proof. (Left for the student.)
236. Corollary 1. To construct a triangle when a side and
two angles are given.
There are two cases to be considered: (1) when the given side is
included between the given angles ; and (2) when it is not (in which
case find the other angle by § 107).
237. Corollary 2. To construct a triangle when the three
sides are given.
238. Corollary 3. To construct a parallelogram when two
sides and the included angle are given.
Combine § 235 and § 233.
134 BOOK II. PLANE GEOMETRY
Proposition XXVIII. Problem
239. To construct a triangle ivheji two sides and the
angle opposite one of them are given.
yY
O/'
Given a and h two sides of a triangle, and A the angle opposite a.
Required to construct the triangle.
Construction. Case 1. If a is less than h.
Construct Z XA Y equal to the given A A. § 232
On A Y take A C equal to h.
Prom C as a center, with a radius equal to a, describe an arc
intersecting the line AX dX B and B\
Draw EC and B'C, thus completing the triangle.
Then both the ^ABC and AB^C satisfy the conditions, and
hence we have two constructions. q.e.f.
This is called the ambiguous case.
Discussion. If the given side a is equal
to the ± CB, the arc described from C will yi
touch AX, and there will be but one con
struction, the rt. A ABC.
If the given side a is less than the per
pendicular from C, the arc described from ^
C will not intersect or touch AX, and hence yY
a construction is impossible. (j /
If Z A is right or obtuse, a construe y^ \ ^
tion is impossible, since a<h\ for the side X \ ^
of a triangle opposite a right or obtuse angle X
is the longest side (§114). A
.Y
PROBLEMS OF CONSTRUCTION 135
Case 2. If a is equal to b.
If the given Z. A is acute, and a = b, the arc described
from C as a center, and with a radius equal to a, will cut
the line WX at the points A and B. y
There is therefore but one triangle that ?<^
satisfies the conditions, namely the isos ^^ :^ll'____ \/ _ v
celes A ABC. ^^^ — ''^
Discussion. If the ZA is right or obtuse, a construction is impossible
when a = b ; for equal sides of a triangle have equal angles opposite them,
and a triangle cannot have two right angles or two obtuse angles (§ 109).
Case 3. If a is greater than h.
If the given Z A is acute, the arc described from C will
cut the line WX on opposite sides of .1, at B and B'. The
A ABC satisfies the conditions, but the A AB'C does not,
for it does not contain the acute Z.A.
C "^
There is then only one triangle that • a^y/^\a ]
satisfies the conditions, namely the ^..l\grS/. ^:>J...x
A ABC. ^'^^< '^^'^
If the given Z .1 is right, the arc described
from C cuts the line WX on opposite sides of A
A at the points B and B', and we have the x^ / j^>\ /
two congruent right triangles ABC and AB'C ^^ "b^"^~~"^'b''^
that satisfy the conditions.
If the given Z .1 is obtuse, the arc de \^
scribed from C cuts the line WX on , n/ \\a ,
oj)posite sides of A, at the points B and ^y i^'l i..i^i.x
B'. The A ABC satisfies the conditions,
but the A AB'C does not, for it does not contain the obtuse
Z.A. There is then only one triangle that satisfies the con
ditions, namely the A ABC,
Discussion. We therefore see that when a > 6, we have only one
triangle that satisfies the conditions, for the two congruent right tri
angles give us only one distinct triangle.
136 BOOK 11. PLANE GEOMETRY
Proposition XXIX. Problem
240. To circumscribe a circle about a given triangle.
^^ ^c
Given the triangle ABC.
Required to circumscribe a O about A ABC.
Construction. Draw the perpendicular bisectors of the sides
^^and^C. §229
Since AB is not the prolongation of CA, these Js will inter
sect at some point 0. Otherwise they would be II, and one of
them would have to be ±. to two intersecting lines. § 82
With as a center, and a radius OA, describe a circle. Post. 4
The O^^C is the O required. q.e.f.
Proof. The point is equidistant from A and B, and also is
equidistant from A and C. § 150
.*. the point is equidistant from A, B, and C.
.*. a O described with as a center, with a radius equal to OA,
will pass through the vertices A, B, and C, by § 160. q.e.d.
241. Corollary 1. To describe a circle through three points
not in the same straight line.
242. Corollary 2. To find the center of a given circle or
of the circle of which an arc is given.
243. Circumcenter. The center of the circle circumscribed
about a polygon is called the circumcenter of the polygon.
PROBLEMS OF CONSTEUCTION
Proposition XXX. Problem
244. To inscribe a circle in a given triangle.
137
P B
Given the triangle ABC.
Required to inscribe a O in A ABC.
Construction. Bisect the A A and B. § 231
From 0, the intersection of the bisectors, draw OP 1. to the
side AB. § 227
With as a center and a radius OP, describe the O PQR.
The OPQR is the O required. q.e.f.
Proof. Since is ii> the bisector of the Z.A,it is equidistant
from the sides AB and AC; and since is in the bisector of
the Z.B, it is equidistant from the sides AB and BC. § 152
.'.a circle described with as a center, and a radius OP,
will touch the sides of the triangle, by § 184. q.e.d.
245. Incenters and Excenters. The center of a circle inscribed
in a polygon is called the incenter of the polygon.
The intersections of the bisectors
of the exterior angles of a triangle
are the centers of three circles, each
tangent to one side of the triangle
and the two other sides produced.
These three circles are called escribed
circles ; and their centers are called
the excenters of the triangle.
138
BOOK II. PLANE GEOMETRY
Proposition XXXI. Problem
246. Through a given. point, to draw a tangent to a
given circle.
. — .P
^ M,'
^''.P
Fig. 1
Given the point P and the circle with center 0.
Required through P to draw a tangent to the circle.
Case 1. When the given point is on the circle (Fig. 1).
Construction. From the center draw the radius OP. Post. 1
Through P draw AT _L to OP. § 228
Then AF is the tangent required. q.e.f.
Proof. Since AF is ± to the radius OP, Const.
.*. AF is tangent to the O at P, by § 184. q.e.d.
Case 2. When the given point is outside the circle (Fig. 2).
Construction. . Draw OP. Post. 1
Bisect OP. § 229
With the midpoint of OP as a center and a radius equal to
I" OP, describe a circle intersecting the given circle at the
points M and N, and draw PM.
Then PM is the tangent required. Q. e. f.
Proof. Draw OM.
Z OMP is a right angle. § 215
.'.PilfisJ to OJ/. !27
.*. PM is tangent to the circle at M, by § 184. q.e.d.
Discussion. In like manner, we may prove PN tangent to the given O.
PEOBLEMS OF CONSTRUCTION 139
Proposition XXXII. Problem
247. Upon a given line as a chord, to describe a segment
of a circle in which a given angle may he inscribed.
,' /o \ \ /
\ / ! ""v '. ly
Given the line AB and the angle m.
Required on AB as a chords to describe a segment of a circle
in which Z. m may be inscribed.
Construction. Construct the ZABX equal to the Z.m. § 232
Bisect the line AB by the ± PO. § 229
From the point B draw BO _L to XB. § 228
• With 0, the point of intersection of PO and BO, as a center,
and a radius equal to OB, describe a circle.
The segment ABQ is the segment required. q.e.f.
Proof. The point is equidistant from A and B. § 150
.'. the circle will pass through A and B. § 160
But BX is ± to OB. Const.
.'. BX is tangent to the O. § 184
.'. AABX is measured by ^ arc ^5. § 220
But any angle, as the Z Q, inscribed in the segment ABQ is
measured by ^ a,vcAB. § 214
.\ZQ = ZABX. Ax. 8
But Z.ABX = Zm. Const.
.'. Z.m may be inscribed in the segment ABQ, by § 217. Q. e.d.
140 BOOK 11. PLANE GEOMETRY
248. How to attack a Problem. There are three common
methods by which to attack a new problem :
(1) By synthesis ;
(2) By analysis ;
(3) By the intersection of loci.
249. Synthetic Method. If a problem is so simple that the
solution is obvious from a known proposition, we have only to
make the construction according to the proposition, and then
to give the synthetic proof, if a proof is necessary, that the
construction is correct.
It is rarely the case, however, that a problem is so simple as to allow
this method to be used. We therefore commonly resort at once to the
second method.
250. Analytic Method. This is the usual method of attack,
and is as follows :
(1) Suppose the problem solved and see what results follow.
(2) Then see if it is possible to attain these results and thus
effect the required construction; in other words, try to work
backwards.
The third method, by the intersection of loci, is considered on page 143.
251. Determinate, Indeterminate, and Impossible Cases. A
problem that has a definite number of solutions is said to be
determinate. A problem that has an indefinite number of solu
tions is said to be indeterminate. A problem that has no solu
tion is said to be impossible.
For example, to construct a triangle, having given its sides, is deter
minate ; to construct a quadrilateral, having given its sides, is indetermi
nate ; to construct a triangle with sides 2 in., 3 in., and 6 in. is impossible.
252. Discussion. The examination of a problem with refer
ence to all possible conditions, particularly with respect to the
number of solutions, is called the discussion of the problem.
Discussions have been given in several of the preceding problems.
SOLUTION OF PKOBLEMS
141
253. Applications of the Anal3rtic Method. The following are
examples of the use of analysis in the solution of problems.
EXERCISE 37
1. In a triangle ABC, to draw PQ parallel to the base AB,
cutting the sides in P and Q, so that PQ shall equal AP \ BQ.
Analysis. Assume the problem solved.
Then AP must equal some part of PQ, as PX,
and BQ must equal QX.
But if AP = PX, what must Z PXA equal ?
•.• PQ is II to AB, what does Z PXA equal ?
Then why must ZBAX = ZXAP?
Similarly, what about ZQBX and Z XBA ?
Construction. Now reverse the process. "What should we do to zi yl and
B in order to fix X ? Then how shall PQ be drawn ? Now give the proof.
2. To construct a triangle, having given the perimeter, one
angle, and the altitude from the vertex of the given angle.
Analysis. Let ABC hQ the triangle, Z G the given angle, and CP
the given altitude, and assume that the problem is solved.
Since the perimeter is
given as a definite line, we X
now try producing AB and
BA, making BN=BC, and
AM=AC.
Then Zm = what angle,
and Zn = what angle ?
Then Zm\ Zn\ Z MCN = 180°.
But ZMCN=Zm'^ZACB\Zn\
.'. 2 Zm + 2 Zn^ZACB = 180°. (Why?)
.. Zm + Zn + IZACB = 90°,
or Zm{ Zn = 90°iZACB.
.: Z MCN =90° \^ZA CB. (Why ?)
.. Z MCN is known.
Construction. Now reverse the process. Draw MN equal to the perim
eter. Then on MN construct a segment in which Z MCN may be inscribed
(§ 247). Draw XC II to MN at the distance CP from MN, cutting the arc
at C. Then A and B are on the ± bisectors of CM and CN. Why ?
PB
142 BOOK II. PLANE GEOMETRY
3. To draw through two sides of a triangle a line parallel
to the third side, so that the part intercepted c ^
between the sides shall have a given length.
If PQ = d, what does AR equal ? How will you
reverse the reasoning ? A R ~b
4. To draw a tangent to a given circle so that it shall be
parallel to a given line.
5. To construct a triangle, having given a side, an adjacent
angle, and the difference of the other sides.
If AB^ ZA, and AC — BC are known, what points are
determined ? Then can XB be drawn ? What kind of a tri ^
angle is A XBC ? How can C be located ? ^ «
6. To construct a triangle, having given two angles and
the sum of two sides. ^
Can the third Z be found ? Assume the prob
lem solved. If AX = AB + BC, what kind of a
triangle is A BXC ? What does Z CBA equal ?
Is Z X known ? How can C be fixed ?
7. To construct a square, having given the diagonal.
8. To draw through a given point P between the sides of
an angle AOB ?i line terminated by the sides ^
of the angle and bisected at P. /
If PM=: PN, and PR is II to AO, what can you 2?/>_>p
say as to OR and RN? Can you now reverse this? /
Similarly, if PQ is II to BO, is OQ = to QM? ^ ^ MA
9. To draw a line that would bisect the angle formed by
two lines if those lines were produced to meet.
li AB and CD are the given lines, consider what would be the con
ditions if they could be produced to meet at 0. Then the q
bisector of Z O would be the ± bisector of PQ, a line drawn /\s^
so as to make equal angles with the two given lines. / I \
Now reverse this. How can we draw PQ so as to make / j >^
ZP = ZQ? Draw BR II to DC, and lay off BR = BQ. ^
Then draw QRP and prove that this is such a line. Then
draw its _L bisector.
EXERCISES IN LOCI 14B
254. Intersection of Loci. The third general method of attack
mentioned in § 248 is by intersection of loci. This is very con
venient when we wish to find a point satisfying two conditions,
each of which involves some locus.
EXERCISE 38
1. To find a point that is i in. from a given point and ^\ in.
from a given line. ^'— _w_
If P is the given point, what is the locus of a — I ; ^ B
a point I in. from P ? If ^^ is the given line, > ^ l__
what is the locus of a point y\ in. from AB ? —■'
These two loci intersect in how many points at most? Discuss the
solution.
2. To find a point that is J in. from one given point and  in.
from another given point.
Discuss the number of possible points answering the conditions.
3. To find a point that is \ in. from the vertex of an angle
and equidistant from the sides of the angle.
4. To find a point that is equidistant from two intersecting
lines and \ in. from their point of intersection.
How many such points can always be found ?
5. To find a point that is \ in. from a given point and equi
distant from two intersecting lines.
Discuss the problem for various positions of the given point.
6. To find a point that is \ in. from a given point and equi
distant from two parallel lines.
Discuss the problem for various positions of the given point.
7. Find the locus of the midpoint of a chord of a given
length that can be drawn in a given circle.
8. rind the locus of the midpoint of a chord drawn through
a given point within a given circle.
Ul G
'0\
144 BOOK II. PLANE GEOMETRY
9. To describe a circle that shall pass through a given point
and cut equal chords of a given length from two parallels.
Analysis. Let A be the given point, BC and DE the given parallels,
MN the given length, and O the center of the required circle.
Since the circle cuts equal chords from tw^o
parallels, what must be the relative distance b
of its center from each ? Therefore what line
must be one locus for ? F
Draw the ± bisector of MN^ cutting F(? at P.
How, then, does VM compare with the radius b~M
of the circle required ? How shall we then find
a point O on FG that is at a distance TM from J. ? Do we then know
that is the center of the required circle ?
10. To describe a circle that shall be tangent to each of two
given intersecting lines.
11. To find in a given line a point that is equidistant from
tAvo given points.
12. To find a point that is equidistant from two \\ "^^
given points and at a given distance from a third p \x' q
given point.
13. To describe a circle that has a given radius and passes
through two given points.
14. To find a point at given distances from two given points.
15. To describe a circle that has its center in a given line
and passes through two given points.
16. To find a point that is equidistant from two given points
and also equidistant from two given intersecting lines.
17. To find a point that is equidistant from two given points
and also equidistant from two given parallel lines. ,
18. To find a point that is equidistant from c A'~~^, ^
two given intersecting lines and at a given dis
tance from a given point.
19. To find a point that lies in one side of '^ '^
a given triangle and is equidistant from the other two sides.
EXEECISES
145
255. General Directions for solving Problems. In attacking a
new problem draw the most general figure possible and the
solution may be evident at once. If the solution is not evident,
see if it depends on finding a point, in which case see if two
loci can be found. If this is not the case, assume the problem
solved and try to work backwards, — the method of analysis.
EXERCISE 39
1. To draw a common tangent to two given circles.
If the centers are and 0' and the radii r and r^, the tangent QR
seems to be II to CKJlf, a tangent from 0' to a circle whose radius is r — r'.
If this is true, we can easily reverse the process. Since there are two
tangents from (X, so there are two common tangents.
In the righthand figure tlie tangent QR seems to be II to (XM, a tangent
from CK to a circle whose radius is r + r'. If this is true, we can easily
reverse the process. There are four common tangents in general.
2. To draw a common tangent to two given circles, using the
following figures.
3. The locus of the vertex of a right triangle, having a
given hypotenuse as its base, is the circle described upon the
given hypotenuse as a diameter.
4. The locus of the vertex of a triangle, having a given base
and a given angle at the vertex, is the arc which forms with
the base a segment in which the given angle may be inscribed.
146
BOOK II. PLANE GEOMETEY
To construct an isosceles triangle^ having given :
5. The base and the angle at the vertex.
6. The base and the radius of the circumscribed circle.
7. The base and the radius of the inscribed circle.
8. The perimeter and the altitude. c
Let ABC be the A required, EF the given
perimeter. The altitude CD passes through the ^^
^i
middle of EF, and the A EA C, BFC are isosceles. A D B
To construct a right triangle, having given :
9. The hypotenuse and one side.
10. One side and the altitude upon the hypotenuse.
11. The median and the altitude upon the hypotenuse. 
12. The hypotenuse and the altitude upon the hypotenuse.
13. The radius of the inscribed circle and one side.
14. The radius of the inscribed circle and an acute angle.
To construct a triangle, having given :
15. The base, the altitude, and an angle at the base.
16. The base, the altitude, and the angle at the vertex.
1 7. One side, an adjacent angle, and the sum of the other sides.
18. To construct an equilateral triangle, hav , — .c
ing given the radius of the circumscribed circle.
19. To construct a rectangle, having given one
side and the angle between the diagonals.
20. Given two perpendiculars, AB and CD, ^
intersecting in 0, and a line intersecting
these perpendiculars in E and F; to con ^
struct a square, one of whose angles shall ^ q
coincide with one of the right angles at O,
and the vertex of the opposite angle of the
square shall lie in EF. (Two solutions.)
A^
. o
r)iB
EXERCISES
147
/«
21. A straight rod moves so that its ends con
stantly touch two fixed rods perpendicular to
each other. Find the locus of its midpoint.
22. A line moves so that it remains par
allel to a given line, and so that one end
lies on a given circle. Find the locus of the
other end.
23. Find the locus of the mid
point of a linesegment that is drawn
from a given external point to a given
circle.
24. To draw lines from two given points
P and Q which shall meet on a given line
AB and make equal angles with AB.
'.' Z BEQ = Z PEC, .'. Z CEP' = Z PEC. (Why ?)
But it is easy to make Z CEP'= Z PEC, by mak
ing PP' JlAB, and CP' = PC, and joining P' and Q.
25. To find the shortest path from a point P to a line AB
and thence to a point Q. q
Prove that PE + EQ<PF\ FQ, where Z BEQ
= Z PEC. _i
This shows that a ray of light from a point to a ^ €■
plane mirror and thence to another jwint takes the i'''
shortest possible path.
26. The bisectors of the angles included by the opposite
sides (produced) of an inscribed quadrilateral intersect at
right angles.
Arc ^X— arc MD
= arc XB 
Arc YA  arc BN
= arc DY
.'. arc YX + arc NM
= arc MY + arc XN. (Why ?)
.. Z YIX = Z XIN. (Why ?)
How does this prove the proposition ? Discuss the impossible case.
P
E^' F B
arc CM. (Why ?)
arc NC. (Why ?)
148 BOOK II. PLANE GEOMETRY
27. Construct this design, making the figure
twice this size.
Construct the equilateral A. Then describe the small
(D with half the side of the A as a radius. Then find
the radius of the circumscribing O.
28. A circular window in a church has a de
sign similar to the accompanying figure. Draw
it, making the figure twice this size.
This is made from the figure of the preceding exer
cise, by erasing certain lines.
29. Two wheels of radii 1 ft. 6 in. and 2 ft. 3 in. respec
tively are connected by a belt, drawn straight between the
points of tangency. The centers being 6 ft. apart, draw the
figure mathematically. Use the scale of 1 in. to the foot.
30. A water wheel is broken and all but a fragment is lost.
A workman wishes to restore the wheel. Make
drawing showing how he can construct a wheel
the size of the original.
31. In this figure Zm = 62°, ^ndZn=2S°.
Find the number of degrees in each of the
other angles, and determine whether ^^ is
a diameter.
32. In this figure ZB = 4.1°, ZA = 65°,
and ZBDC = 97°. Eind the number of
degrees in each of the other angles, and
determine whether CD is a diameter.
33. Construct or explain why it is im ^' ~
possible to construct a triangle with sides 3 in., 2 in., 6 in. ;
also one with sides 5 in., 7 in., 12 in.; also one
with sides 2 in., 1 in., 1^ in.
34. Show how to draw a tangent to this circle p
at the point P, the center of the circle not being
accessible.
.1^^^
EXERCISES 149
EXERCISE 40
1. In a circle whose center is O the chord AB is drawn so
that Z BA O = 27°. How many degrees are there in Z.AOB?
2. In a circle whose center is the chord AB is drawn so
that Z.BAO = 25°. On the circle, and on the same side of AB
as the center 0, the point D is taken and is joined to A and B.
How many degrees are there in Z.ADB?
3. What is the locus of the midpoint of a chord of a circle
formed by secants drawn from a given external point ?
4. In a circle whose center is two perpendiculars OM and
ON are drawn to the chords AB and CD respectively, and it
is known that Z NMO = Z ONM. Prove that AB=CD.
5. Two circles intersect at the points A and B. Through
A a variable secant is drawn, cutting the circles at C and D.
Prove that the angle DBC is constant.
6. Let A and B be two fixed points on a given circle, and M
and N be the extremities of a rotating diameter of the same
circle. Find the locus of the point of intersection of the
lines AM and BN.
7. Upon a line AB a, segment of a circle containing 240° is
constructed, and in the segment any chord PQ subtending an
arc of 60° is drawn. Find the locus of the point of intersection
of AP and BQ ; also of A Q and BP.
8. To construct a square, given the sum of the diagonal and
one side. ^d
Let A BCD be the square required, and CA the di j^ ^ A/ \
agonal. Produce CA, making AE = AB. A ABC and """>.,:;, /'
ABE are isosceles and ABAC = ZACB = i6°. Find B
the value of Z E. Construct Z CBE. Now reverse the reasoning.
The propositions in Exercise 40 are taken from recent college entrance
examination papers.
150 BOOK II. PLANE GEOMETEY
EXERCISE 41
Review Questions
1. Define the word circle and the principal terms used in
connection with it.
2. What is meant by a central angle ? How is it measured ?
3. What is meant by an inscribed angle? How is it measured?
4. State the general proposition covering all the cases that
have been considered relating to the measure of an angle formed
by the intersection of two secants.
5. State all of the facts you have learned relating to equal
chords of a circle.
6. State all of the facts you have learned relating to unequal
chords of a circle.
7. State all of the facts you have learned relating to tangents
to a circle.
8. How many points are required to determine a straight
line ? two parallel lines ? an angle ? a circle ?
9. Name one kind of magnitude that you have learned to
trisect, and state how you proceed to trisect this magnitude.
10. In order to construct a definite triangle, what parts must
be known ?
11. What are the important methods of attacking a new
problem in geometry ? Which is the best method to try first ?
12. What is meant by determinate, indeterminate, and im
possible cases in the solution of a problem ?
13. Distinguish between a constant and a variable, and give
an illustration of each.
14. Distinguish between inscribed, circumscribed, and escribed
circles.
15. What is meant by the statement that a central angle is
measured by the intercepted arc ?
BOOK III
PROPORTION. SIMILAR POLYGONS
256. Proportion. An expression of equality between two
equal ratios is called a proportion.
257. Symbols. A proportion is written in one of the fol
lowing forms : 7 = ~;5 a:b = c : d\ a : b : : c : d.
This proportion is read " a is to 6 as c is to d " ; or " the ratio of a to 5
is equal to the ratio of c to d."
258. Terms. In a proportion the four quantities compared
are called the terms. The first and third terms are called the
antecedents; the second and fourth terms, the consequents.
The first and fourth terms are called the extremes; the second
and third terms, the means.
Thus in the proportion a:b = c :d, a and c are the antecedents, b
and d the consequents, a and d the extremes, b and c the means.
259. Fourth Proportional. The fourth term of a proportion
is called the fourth proportional to the terms taken in order.
Thus in the proportion a : 6 = c : d, d is the fourth proportional to
a, &, and c.
260. Continued Proportion. The quantities a, b, c, d, • • • are
said to be in continued proportion, if a : b = b : c = c : d = •  • .
If three quantities are in continued proportion, the second
is called the mean proportional between the other two, and the
third is called the third proportional to the other two.
Thus in the proportion a.b = b:c,b is the mean proportional between
a and c, and c is the third proportional to a and b.
151
152 BOOK III. PLANE GEOMETRY
Proposition I. Theorem
261. In any proportion the product of the extremes is
equal to the product of the means.
Given a:b = c:d.
To prove that ad = he.
Proof. j = ^' §257
a
Multiplying by bd, ad = bc, by Ax. 3. " q.e.d.
262. Corollary 1. The mean proportional between tivo
quantities is equal to the square root of their product.
For if a:h = h:c, then h^ = ac (§ 261), and h = Vac, by Ax. 5.
263. Corollary 2. If the two antecedents of a proportion
are equals the two consequents are equal.
264. Corollary 3. If the product of two quantities is equal
to the product of two others^ either two may be made the extremes
of a proportion in which the other two are made the means.
For if ad = 6c, then, by dividing by 6d,  = , by Ax. 4.
6 d
Proposition II. Theorem
265. If four quantities are in proportion, they are in
proportion hy alteimation ; that is, the first term is to
the third as the second term is to the fourth.
or
Given
a: b = c:d.
To prove that
a: c = b: d.
Proof.
ad = be.
§261
Dividing by cd,
a b
c = d'
Ax. 4
a:c = b:d,hy % 257.
Q.E.D.
THEORY OF PROPORTION 153
Proposition III. Theorem
266. If four quantities are in proportion, they are in
proportion hy inversion ; that is, the second term is to
the first as the fourth term is to the third.
Given a:b = c:d.
To prove that h: a = d: c.
Proof. bc = ad. §261
Dividing each member of the equation by ac,
 = > Ax. 4
a c
or b : a = d : c, hy ^ 257. Q. e. d.
Proposition IV. Theorem
267. If four quantities are in proportion, they are in
proportion hy composition; that is, the sum of the first
two terms is to the second term as the sum of the last
tivo terms is to the fourth term.
Given a:b=c:d.
To prove that a\h :h = c \ d: d.
Proof. j =  §257
d
Adding 1 to each member of the equation,
a ^ G ^
a\b c\d
b ^ d '
.. a[b:b = c + d:d,hy %257. Q.e.d.
In a similar manner it may be shown that
a { b : a = c \ d : c.
a ^ c ^ . ^
 + 1 = +1, Ax.l
or
154 BOOK III. PLANE GEOMETRY
Proposition V. Theorem
268. If four quantities are in proportion , they are in
proportion hy division; that is, the difference of the
first two terms is to the second term as the difference
of the last two terms is to the fourth term.
§257
Ax. 2
or
Given
a:b = c:d.
To prove that
a — b: h = c — d: d.
Proof.
a c
h~d'
"h ^ d ^'
a — h c — d
h d
.'.ah:h = cd:d,hj §257.
In a similar manner
it may be shown that
a — b:a = c — d:c.
Q.E.D.
Proposition VI. Theorem
269. Li a series of equal ratios, the sum of the ante
cedents is to the su7n of the consequents as any ante
cedent is to its consequent.
Given a:b = c:d=:e:f=g:h.
To prove that a\ e \ e^ g:h \ d +/+ h = a:h.
Proof. Let r =  =  =  = 'z
d J I I
Then a = hr, c = dr, e=fr, g = hr. Ax. 3
,',a.\e + e\g = {h + d+f+h)r. Ax. 1
^ CL \ c \ e \ g a
Ax. 4
"b\d+f+h h
a^c\e + g:b\d +/+ h = a : h, by § 257. Q.E.D.
THEORY OF PROPORTION 155
Proposition VII. Theorem
270. Like powers of the terms of a proportion are iri
proportion.
Given a:bz=c: d.
To prove that a" : 5" = ^ : (^.
Proof. X = ^ ' §257
a
•*• tt^'t:' by Ax. 5. q.e.d.
Proposition VIII. Theorem
271. If three quantities are in continued proportion,
the first is to the third as the square of the first is to
the square of the second.
Given a:b=b:c.
To prove that a: c = a^:b\
Proof.
a^ = a%
Iden.
and
ae = b\
a" a a^
ac c b^
§261
Ax. 4
. • . a :c = a'^:
h%
by §
257.
Q.E.D.
272. Nature of the Quantities in a Proportion. Although we
may have ratios of lines, or of areas, or of solids, or of angles,
we treat all of the terms of a proportion as numbers.
If b and d are lines or solids, for example, we cannot
multiply each member of  =  by bd, as in § 261.
Hence when we speak of the product of two geometric viagni
tudes, we mean the product of the numbers that represent them
when expressed in terms of a common unit.
166. BOOK III. PLANE GEOMETEY
EXERCISE 42
1. Prove that a:h = ma : mfib.
2. \i. a\h=^G\d^ and m : 71=.]) : q, prove that am :bn=cp : dq.
If a:b = c: d, prove the following :
3. a:d = bc:d\ 7. ma : nb = me : nd.
4. l:b = c: ad. S. a — l:b = bc — d:bd.
5. ad:b = c:l. 9. a^l:l = bc + d:d.
6. ma : b = mc : d. 10. 1 : ^c = 1 : ad.
11. a\b:a — b = c\d:c — d.
In Ex. 11, use § 267 and § 268, and Ax. 4. In this case a, 6, c, and d
are said to be in proportion by composition and division.
If a:b = b: c, prove the following :
12. G'.b = b'.a. 14. (^ + V^) (Z>  V^) = 0.
13. a:c = P:c''. 15. ac l:b 1 = b \l'.l.
16. If 2:7= 3: a, show that 2^^ = 21, andic=10^.
^mc? ^Ae value of x in the following :
17. l:7=3:ic. 29. ic : 2.7 = 7: 5.4.
18. 2:9 = 5:ic. 30. cc : 8.1 = 0.3 : 0.9.
19. 4:28 = 3:ic. 31. 2:ic = £c:32.
20. 2:^ = a;:12. 32. 7:£c = cc:28.
21. 3:5 = ;:c:9. 33. l:lfx = a; 1: 3.
22. 7:21=£c:5. 34. 5 :£c  2 = a: + 2 :1.
23. 3:5 = £c 41:10. 35. cc2:2a = 3a:6.
24. 8:15 = 2ic + 3:45. 36. x'.4.a = 2a^:x\
25. 0.8:ic = 4:9. 37. a:l = xl:l.
26. 0.7:a; = 21:15. 38. a; +l:ic 1= 3 : 2.
27. 0.25:cc = 5:8. 39. 3 :cc + 4 = a;  4: 3.
28. a;:1.3 = 4:0.26. 40. ab:b = bcx'.bcx.
PKOPORTIOKAL LINES 157
Proposition IX. Theorem
273. If a line is drawn through two sides of a tri
angle parallel to the third side, it divides the tivo sides
proportionally.
Given the triangle ABC, with EF drawn parallel to BC.
To prove that EB:AE=FC: AF.
Case 1. When AE and EB are commensurable.
Proof. Assume that MB is a common measure oi AE and EB.
Let MB be contained m times in EB, and n times in AE.
Then EB:AE = m:n.
{For m and n are the numerical measures of EB and AE.)
At the points of division on EB and AE draw lines II to BC.
These lines will divide A C into m\n equal parts, of which
FC will contain m parts, and AF will contain n parts. § 134
.•.FC:AF=m:n.
.'.EB:AE = FC:AF,hyAx.S. Q.e.d.
For practical purposes this proves the proposition, for even if AE and
EB are incommensurable, we can, by taking a unit of measure small
enough, find the measure of AE and EB to as close a degree of approxi
mation as we may desire, just as we can carry V2 to as many decimal
places as we wish, although its exact value cannot be expressed rationally.
On this account many teachers omit the incommensurable case dis
cussed on page 158, or merely require the proof there given to be read
aloud and explained by the class.
158 BOOK III. PLANE GEOMETEY
Case 2. When AE and EB are incommensurable,
A
B (J
Proof. Divide AE into a number of equal parts, and apply
one of these parts to EB as many times as EB will contain it.
Since AE and EB are incommensurable, a certain number of
these parts will extend from E to some point G', leaving a
remainder GB less than one of these parts.
Draw GH II to BC.
Then EG:AE = FN : AF. Case 1
By increasing the number of equal parts into which AE is
divided, we can make the length of each part less than any
assigned positive value, however small, but not zero.
Hence GB, which is less than one of these equal parts, has
zero for a limit. § 204
And the corresponding segment HC has zero for a limit.
Therefore EG approaches EB as a limit,
and FH approaches EC as a limit.
.*. the variable — — approaches ~— as a limit,
AJii All/
and the variable — — approaches — — as a limit.
AF ^^ AF
But — — is always equal to — — • Case 1
.•.f = f,by.207.
A
Fl
L
\«
\
\m
I
\\
PKOPORTIONAL LINES 159
274. Corollary 1. One side of a triangle is to either of
its segments cut off hy a line parallel to the base as the third
side is to its corresponding segment.
For EB.AE = FC.AF. §273
By composition, EB + AE: AE = FC + AF: AF, § 267
or AB:AE = AC:AF. Ax. 11
275. Corollary 2. Three or more parallel lines cut off
proportional intercepts on any two transversals.
Draw ^iV II to CD.
Then AL = CG, LM = GK, MN = KB. § 127
Now AH:AM= AF:AL=FH:LM, §274
and AH:AM=HB:MN. §273
.'.AF'.CG =FH:GK=HB:KB. Ax.9 b N D
EXERCISE 43
1. In the figure of § 275, suppose AH = 5 in., AF=2 in.,
and CK=6 in. Find the length of CG. d c
2. In this square PQ is II to AB. If a side of the p
square is 10 in., Z)^ = 14.14 in. If DP = 3 in.,
what is the length ot BQ?
3. The sides of a triangle are respectively 3 in., 4 in., and
5 in. A line is drawn parallel to the 4inch side, cutting the
3inch side 1 in. from the vertex of the largest angle. Find
the length of the two segments cut from the longest side.
4. Two pieces of timber 1 ft. wide are fitted together at
right angles as here shown. AB is 8 ft. long, AC 6 ft. long,
and the distance BCj along the dotted line,
is 10 ft. A carpenter finds it necessary to
saw along the dotted line. Find the length
of the slanting cut across the upright piece ;
across the horizontal piece.
160 BOOK III. PLANE GEOMETEY
Proposition X. Theorem
276. If a line divides two sides of a triangle jpro^or
tionally, it is parallel to the third side.
or
B G
Given the triangle ABC with EF drawn so that
EBFC
AE~ AF'
To prove that EF is II to BC.
Proof. Suppose that EF is not parallel to BC.
Then from E draw some other line, as EH, parallel
Then AB : AE = AC : AH.
{One side of a A is to either of its segments cut off by a line II
base as the third side is to its corresponding segment.)
But EB:AE = FC:AF.
.. EB + AE: AE = EC \ AFiAF,
AB:AE = AC:AF.
.'.AC:AF=AC:AH.
.\AF=AH.
{For the two antecedents are equal.)
.'. EF and EH must coincide.
{For their end points coincide.)
EH is II to BC.
.'. EF, which coincides with EH, is II to BC.
This proposition is the converse of Prop. IX.
But
to^C.
§274
to the
Given
§267
Ax. 11
Ax. 8
§263
Post. 1
Const.
Q.E.D.
PKOPOETIONAL LINES 161
277. Dividing a Line into Segments. If a given line AB is
divided at P, a point between the extremities A and B, it is
said to be divided internally into the segments AP and PB'j
and if it is divided at P', a point in the prolongation of BA, it is
said to be divided externally into the segments AP' and P'B.
P' A P B
In either case the length of the segment is the distance from the
point of division to an extremity of the line. If the line is divided
internally, the sum of the segments is equal to the line ; and if the line
is divided externally, the difference of the segments is equal to the line.
Suppose it is required to divide the given line AB internally
and externally in the same ratio ; as, for example, in the ratio
of the two numbers 3 and 5.
P' APE
We divide AB into 3 + 5, or 8, equal parts, and take 3 parts
from A ; we then have the point P, such that
AP'.PB = Z:n. (1)
Secondly, we divide AB into 5 — 3, or 2, equal parts, and lay
off on the prolongation of BA three of these equal parts ; we
then have the point P', such that
ylP':P'5 = 3:5. (2)
Comparing (1) and (2), we have
AP:PB = AP''.P'B.
278. Harmonic Division. If a given straight line is divided
internally and externally into segments having the same ratio,
the line is said to be divided harmonically.
Thus the line AB has just been divided internally and externally in
the same ratio, .3 : 5, and ^ JB is therefore said to be divided harmonically
at P and P' in the ratio 3:5.
162 BOOK III. PLANE GEOMETRY
Proposition XI. Theorem
279. The bisector of an angle of a triangle divides
the opposite side into segmeiits which are proportional
to the adjacent sides.
'^G
Given the bisector of the angle C of the triangle ABC^ meeting
ABatM,
To prove that AM: MB = CA: CB.
Proof. Erom A draw a line II to MC.
This line must meet BC produced, because BC and MC
cannot both be parallel to the same line. § 94
Let this line meet BC produced at E.
Then AM:MB=EC :CB. §273
{If a line is drawn through two sides of a A parallel to the third side, it
divides the two sides proportionally.)
Also ZACM=ZCAE, § 100
{Alt.int. A of II lines are equal.)
and Z MCB = ZAEC. § 102
{Ext.int. A of II lines are equal.)
But ZACM=ZMCB. Given
,'.ZCAE = ZAEC. Ax. 8
.\EC=CA. §76
Put CA for its equal ^C in the first proportion.
Then AM:MB= CA :CB,hy Ax. 9. Q. E. D.
PEOPOETIONAL LINES 163
Pboposition XII. Theorem
280. The bisector of an exterior angle of a triangle
divides the opposite side externally into segments ivhich
are proportional to the adjacent sides.
A ^
Given the bisector of the exterior angle ECA of the triangle ABCy
meeting BA produced at M'.
To prove that AM' : M'B = CA:CB.
Proof. Draw AF II to M'C, meeting EC at F.
Then AM': M'B=FC .CB. §274
{One side of a A is to either of its segments cut off by a line II to
the base as the third side is to its corresponding segment.)
Now Z ECM'= Z CFA, § 102
and ZM'CA=ZFAC. §100
But ZECM'=ZM'CA. Given
.'.ZCFA=ZFAC. Ax. 8
.. CA = FC. § 76
Put CA for its equal FC in the first proportion.
Then AM': M'B = CA : CB, by Ax. 9. Q.e.d.
Discussion. In case CA = CB, what is the arrangement of the lines ?
281. Corollary. The bisectors of the interior angle arid the
exterior angle at the same vertex of a triangle, meeting the
opposite side, divide that side harmonically.
164 BOOK III. PLANE GEOMETRY
EXERCISE 44
1. In a triangle ABC, AB = 6.5, CA = 6,BC = 7. Find the
segments oi AB made by the bisector of the angle C
2. In a triangle ABC, CA = 7.5, BC = 7,AB = 8. Find the
segments of CA made by the bisector of the angle B.
3. The sides of a triangle are 12, 16, 20. Find the segments
of the sides made by bisecting the angles.
4. If a spider, in making its web,
makes A'B' II to AB, B'C II to BC, C'D'
II to CD, D'E' II to DE, and E'F' II to EF,
and then runs a line from F' II to FA,
will it strike the point A ' ? Prove it.
5. From any point O within the triangle ABC the lines
OA, OB, OC are drawn and are bisected respectively by A', B',
and C. Prove that Z CBA = Z C'B'A'.
6. Prove Ex. 5 if the point is outside the triangle.
7. From any point within the quadrilateral ABCD lines are
drawn to the vertices A,B,C, D, and are bisected by A',B',C', D'.
Prove that Z CBA = Z C'B'A '.
8. If a pendulum swings at the point 0, cutting two paral
lel lines at P and Q respectively, the ratio OP : OQ is constant.
9. Through a fixed point P a line is drawn cutting a fixed
line at X. PX is then divided at Y so that the ratio PYiYX
is constant. Find the locus of the point F as Z moves along
the fixed line.
10. From the point P on the side CA of the triangle ABC
parallels to the other sides are drawn meeting ABin Q and BC
in R. Prove that AQ: QB = BR : RC.
11. In the triangle ABC, P and Q are taken on the sides CA
and BC so that AP : PC = BQ: QC. AR is then drawn parallel
to PB, meeting CB produced in R. Prove that CB is the mean
proportional between CQ and CR.
SIMILAR POLYGONS
165
282. Similar Polygons. Polygons that have their correspond
ing angles equal, and their corresponding sides proportional,
are called similar polygons.
Thus the polygons ABODE and A'B'C'iyE' are similar, if the A A, B,
C, D, E are equal respectively to the A A% B% C", i/, E\ and if
AB : A'B' = BC : B'C = CD : C'XK = DE : D'E' = EA : E'A\
Similar polygons are commonly said to be of the same shape.
283. Corresponding Lines. In similar polygons those lines
that are similarly situated with respect to the equal angles
are called corresponding lines.
Corresponding lines are also called homologous lines.
284. Ratio of Similitude. The ratio of any two corresponding
lines in similar polygons is called the ratio of similitude of
the polygons.
The primary idea of similarity is likeness of form. The two
conditions necessary for similarity are :
1. For every angle in one of the figures there must he an
equal angle in the other.
2. The corresponding sides must he proportional.
Thus Q and Q! are not similar ; the corresponding sides are
proportional, but the corresponding angles are not equal. Also
R and it' are not similar; the corresponding angles are equal,
but the corresponding sides are not proportional.
Q'
In the case of triangles either condition implies the other.
166 BOOK III. PLANE GEOMETRY
Proposition XIII. Theorem
285. Two rnutually equiangular triangles are similar.
Given the triangles ABC and A'B^C\ having the angles A, B^ C
equal to the angles A', B\ C respectively.
To prove that the A ABC and A'B'C' are similar..
Proof. Since the A are mutually equiangular, Given
we have only to prove that
AB : A^B' = AC : A'C = BC : B'C. § 282
Place the A^'^'C*' on the A^^C so thatZC shall coincide
with its equal, the Z C, and A'B' take the position PQ. Post. 5
Then Aj^ = ^ A. Given
.. PQ is II to AB. § 103
.'.AC:PC = BC: QC ; §274
that is, AC :A'C' = BC iB'C. Ax. 9
Similarly, by placing the A A'B'C on the A ABC so that Z B'
shall coincide with its equal, the Z B, we can prove that
AB:A'B' = BC:B'C'.
.\ AB : A'B' = AC : A'C = BC : B'C. Ax. 8
.. A ABC is similar to A A'B'C, by § 282. q.e.d.
286. Corollary 1. Two triangles are similar if two angles
of the one are equal respectively to two angles of the other.
287. Corollary 2. Two right triangles are similar if an
acute angle of the one is equal to an acute angle of the other.
SIMILAR POLYGONS
167
Proposition XIV. Theorem
288. If two triangles have an angle of the one equal
to an angle of the other ^ and the including sides propor
tional, they are similar.
Given the triangles ABC and A'B'C\ with the angle C equal to
the angle C and with CA : C'A' = CB : C^B\
To prove that the A ABC a7id A'B'C are similar.
Proof. Place the A A'B'C on the A ABC so that Z.C' shall
coincide with its equal, the Z C. Post. 5
Then the A A'B'C will take the position of the A PQC.
CA _ CB ,
CA'~C^''
CA CB
CP~ CQ
CACP CB — CQ
Now
that is,
Given
Ax. 9
or
CP
PA
CP
..'. PQ is
CQ
QB
CQ
to AB.
{If a line divides two sides of a A proportionally, it is
.'. Ap =z Z.A, and Aq = Z.B.
Now /LC = AC'.
.'. A PQC is similar to A ABC.
.'. A A'B'C is similar to A ABC.
268
§276
to the third side.)
§102
Given
§285
Q. E. D.
168
BOOK III. PLANE GEOMETRY
Proposition XY. Theorem
289. If two triangles have their sides respectively
proportional^ they are similar.
A B A' B'
Given the triangles ABC and A'B'C\ having
AB : A'B' =BC:B'C'=CA: CA \
To prove that the A ABC and A'B'C are similar.
Proof. Upon CA take CP equal to CA\ and upon CB take
CQ equal to C'^'; and draw PQ.
Now
Or, since
Also
CA : C'A' = BC:B'C'.
CP = CA', and CQ = C'B',
CA: CP = CB:CQ.
ZC = ZC.
A ABC and,PQC are similar.
Given
Const.
Ax. 9
Iden.
§288
{If two A have an angle of the one equal to an angle of the other, and
the including sides proportional, they are similar.)
that is,
But
But
.\CA:CP = AB:PQ;
CA : CA' = AB : PQ.
CA:CA' = AB:A'B'.
.. AB:PQ = AB:A'B'.
.\PQ = A'B'.
Hence the A PQC and A'B'C Sive congruent.
APQC has been proved similar to A ABC.
.'.A A'B'C is similar to A ABC
§282
Ax. 9
Given
Ax. 8
§263
§80
Q.E.D.
SIMILAR POLYGOJ^S
Proposition XVI. Theorem
169
290. TiDo triangles ivhich have their sides respectively
jjarallel, or ^respectively perpendicular, are similar.
Given the triangles ABC and A'B'C'j with their sides respec
tively parallel ; and the triangles DEF and D'E'F'y with their sides
respectively perpendicular.
To prove that 1. the A ABC and A'B'C are similar ;
2. the A DEF and D'E'F' are similar.
Proof. 1. Produce EC and AC to B'A', forming A x and y.
Then Z.B = Zx(^ 100), Siud ZB' = Zx. §102
.\ZB = ZB'. Ax. 8
In like manner, ZA = ZA'.
.. A ABC is similar to A A'B'C. § 286
2. Produce DE and FD to meet D'E' and F'D' at P and R.
The quadrilateral E'QEP has zip and q right angles. Given
.*. A E' and PEQ are supplementary. § 144
But A y and Pi?Q are supplementary. § 43
Therefore Zy = ZE'. § 58
In like manner, Zx = ZD'.
..A X>£:f is similar to A D'E'F', by § 286. Q.e.d.
Discussion. The parallel sides and the perpendicular sides respectively
are corresponding sides of the triangles.
170 BOOK III. PLANE GEOMETRY
Pkoposition XVII. Theorem
291. The perimeters of two similar polygons have the
same 7'atio as any two corresponding sides.
D
Given the two similar polygons ABCDE and A'B'C'D'E', with
p and />' representing their respective perimeters.
To prove that p ip' = AB : A'B'.
AB _ BC_ _ CD_ _ DE _ EA_
A'B' ~ B'C ~ CD' ~ D'E' ~ EA'' ^
AB { BC { CD \ DE { EA _ AB
A 'B' + B'C + C"X>'+ D'E' + E'A'~ A 'B''
§269
r,2):p' = AB:A'B'jhyAx.9. q.e.d.
EXERCISE 45
1. The corresponding altitudes of two similar triangles have
the same ratio as any two corresponding sides.
2. The base and altitude of a triangle are 15 in. and 7 in.
respectively. The corresponding base of a similar triangle is
3.75 in. Find the corresponding altitude.
3. If two parallels are cut by three concurrent transversals,
the corresponding segments of the parallels are proportional.
4. The point P is any point on the side OX of the angle
XOY. From P a perpendicular PQ is let fall on OY. Prove
that for any position of P on OX the ratio OP : PQ is constant,
and the ratio PQ: OQ is constant.
SIMILAR POLYGONS
171
5. In drawing a map of a triangular field with sides 75 rd.,
60 rd., and 50 rd. respectively, the longest side is drawn 1 in.
long. How long are the other two sides drawn ?
6. This figure represents part of a diagonal scale used by
draftsmen. The distance from to 10
is 1 centimeter, or 10 millimeters. Show
how to measure 5 mm, ; 1 mm. ; 0.9 mm. ;
0.5 mm. ; 1.5 mm. On what proposition
does this depend ? _. _ .
7. This figure represents a pair of proportional
compasses used by draftsmen. By adjusting the
screw at 0, the lengths OA and OC, and the corre
sponding lengths OB and OD, may be varied pro
portionally. Prove that AOAB is always similar
to AOCD. If 0^ = 3 in. and OC = 5 in., then AB
is what part of CD?
8. ABCD is any polygon and P is any
point. On ^ P any point A ' is taken and A 'B'
is drawn parallel to AB as shown. Then
B'C'&nd CD' are drawn parallel to BC and
CD. Is D'A' parallel to DA ? Is A'B'C'D'
similar to ABCD? Prove it.
9. If two circles are tangent externally, the corresponding
segments of two lines drawn through the point of contact and
terminated by the circles are proportional.
10. If two circles are tangent externally, their common ex
ternal tangent is the mean proportional between their diameters.
11. AB and AC are chords drawn from any point /I on
a circle, and AD is ^ diameter. If the tangent at D intersects
AB and A C at i? and F, the triangles ABC and A EF are similar.
12.' If AD and BE are two altitudes of the triangle ABC^ the
triangles DEC and ABC are similar.
p^:
172
BOOK III. PLANE GEOMETRY
Proposition XVIII. Theorem
292. If tivo polygons are similar, they can he sepa
rated hito the same number of triangles, similar each to
each, and similarly placed.
Given two similar polygons ABCDE and A'B'CD'E^ with angles
A, B, C, 2>, E equal to angles ^', 5', C, D\ E' respectively.
To prove that ABODE and A'B'C'D'E' can be separated
into the same number of triangles, similar each to each, and
similarly placed.
Proof. Draw the corresponding diagonals DA, D'A', and
DB, D'B'.
Since Z.E = Z.E\
§282
and
DE:D'E' = EA : E'A',
§282
.'. ADEA and D'E'A' are similar.
§288
In like manner,
ADBC and D'B'C are similar.
Purthermore
ZBAE = ZB'A'E',
§282
and
ZDAE = ZDA'E'.
§282
By subtracting,
ZBAD = ZBA'D'.
Ax. 2
Now
DA :DA' = EA : E'A',
§282
and
AB:A'B' = EA : E'A'.
§282
.
'. DA:D'A' = AB:A'B'.
Ax. 8
..A DAB and D'A'B' are similar.
by § 288.
Q.E.D.
SIMILAR POLYGONS
173
Proposition XIX. Theorem
293. If tivo polygons are composed of the same num
ber of triangles, similar each to each, and similarly
pjlaced, the polygons are similar.
A B A' B'
Given two polygons ABCDE and A'B^C'D'E' composed of the
triangles DEA^ DAB^ DBC, similar respectively to the triangles
D'E'A\ D'A'B', D'B'C, and similarly placed.
To prove that ABCDE is mnilar to A'B'C'D'U'.
Proof. ZE = ZE'.
Also ZDAE = Z D'A 'E'j
and ZBAD = ZB'A'D'.
By adding, ZBAE = ZB'A 'E'.
Similarly Z CBA = Z C'B'A ', and Z EDC = Z E'D'C.
Again, ZC = ZC'. § 282
Hence the polygons are mutually equiangular.
DE _ EA DA AB DB BC CD
~ E'A'
§282
§282
Ax. 1
Also
282
D'E' E'A' D'A' A'B' D'B' B'C CD'
Hence the polygons are not only mutually equiangular but
they have their corresponding sides proportional.
Therefore the polygons are similar, by § 282. q.e.d.
This proposition is the converse of Prop. XVIII.
174 BOOK III. PLANE GEOMETRY
Proposition XX. Theorem
294. If m a right triangle a perpeiidicular is clraivn
from the vertex of the right aiigle to the hypotenuse :
1. The triangles thus formed are similar to the given
triangle, and are similar to each other.
2. The perpendicidar is the mean proportional he
tween the segments of the hypotenuse.
3. Each of the other sides is the mean proportional
hetiveen the hypotenuse and the segment of the hypote
nuse adjacent to that side,
c
A F B
Given the right triangle ABC^ with CF drawn from the vertex
of the right angle C, perpendicular to AB.
1. To prove that the A BCA, CFA, BFC are similar.
Proof. Since the Z. a^ is common to the rt. A CFA and BCA,
.'. these A are similar. §287
Since the Z Z> is common to the rt. A BFC and BCA,
.'. these A are similar. § 287
Since the A CFA and BFC are each similar to A BCA,
.'. these A are mutually equiangular, § 282
Therefore the A CFA and BFC are similar, by § 285. Q. e. d.
2. To prove that AF :CF=CF: FB.
Proof. In the similar A CFA and BFC,
AF: CF=CF:FB, by § 282. q.e.d.
NUMEKICAL PROPERTIES OF FIGURES 175
3. To prove that AB:AC = AC:AF,
and AB:BC = BC:BF.
Proof. In the similar A EC A and CFA,
AB:AC = AC:AF. §282
In the similar ABC A and BFC,
AB:BC = BC: BF, by § 282. q.e.d.
295. Corollary 1. The squares on the two sides of a right
triangle are proportional to the segments of the hypotenuse
adjacent to those sides. ,
From the proportions in § 294, 3,
AAJ^ = ABx AF, and liC^ = AB x BF. § 261
Hence ^^ AB^AF ^A^ ^^ ,
BC^ AB X BF BF
296. Corollary 2. The square on the hypotenuse and the
squai^e on either side of a right triangle are proportional to
the hypotenuse and the seg7nent of the hypotenuse adjacent
to that side.
Since AB^ ^ AB x AB, Iden.
and, as in § 295, AC^ = AB x AF, § 261
AB^ _ ABx AB _AB ^^ ^
'' AC''~ ABx AF~AF'
297. Corollary 3. The perpendicular from any point on
a circle to a diameter is the mean proportional hehveen the
segments of the diameter. g^
298. Corollary A. If a perpendicular
is drawn from any point on a circle to a
diameter^ the chord from that point to ^
either extremity of the diameter is the mean proportional he
tiveen the diameter and the segment adjacent to that chord.
176 BOOK III. PLANE GEOMETRY
EXERCISE 46
1. The perimeters of two similar polygons are 18 in. and
14 in. If a side of the first is 3 in., find the corresponding side
of the second.
2. In two similar triangles, ABC and A'B'C', AB = 6 in.,
BC = 7 in., CA = S in., and A'B' = 9 in. Eind B'C and C'A'.
3. The corresponding bases of two similar triangles are
11 in. and 13 in. The altitude of the first is 6 in. Find the
corresponding altitude of the second.
4. The perimeter of an equilateral triangle is 51 in. Eind
the side of an equilateral triangle of half the altitude.
5. The sides of a polygon are 2 in., 2^ in., 3 in., 3 in., and
5 in. Eind the perimeter of a similar polygon whose longest
side is 7 in.
6. The perimeter of an isosceles triangle is 13, and the ratio
of one of the equal sides to the base is 1§. Eind the three sides.
7. The perimeter of a rectangle is 48 in., and the ratio of
two of the sides is f . Eind the sides.
8. In drawing a map to the scale y^^oVoo) what length will
represent the sides of a county that is a rectangle 25 mi. long
and 10 mi. wide ? Answer to the nearest tenth of an inch.
9. Two circles touch at P. Through P three lines are
drawn, meeting one circle in A, B, C, and the other in
A\ B', C respectively. Prove the triangles ABC, A'B'C' similar.
10. If two circles are tangent internally, all chords of the
greater circle drawn from the point of contact are divided pro
portionally by the smaller circle.
11. In an inscribed quadrilateral the product of the diagonals
is equal to the sum of the products of the opposite
sides.
Draw D^, making ZEDC = Z ABB. The A ABB and
ECB are similar ; and the A BCB and AEB are similar.
NUMERICAL PROPERTIES OF FIGURES 177
Proposition XXI. Theorem
299. If tivo chords intersect ivithin a circle, the prod
uct of the segments of the one is equal to the product
of the segments of the other.
Given the chords AB and CDy intersecting at P.
To prove that PAxFB=FCx PD.
Proof. Draw A C and BD.
Then since Aa = Z.a\ § 214
(Each is measured by  arc CB.)
and Zc = Zc\ §214
{Each is measured by  arc DA.)
.. the A CPA and BPD are similar. § 286
.. PA :PD=PC:PB. §282
.. PAxPB=PC xPD,hj ^261. Q.E.D.
300. Corollary. If two chords intersect within a circle,
the segments of the one are reciprocally proportional to the
segments of the other.
This means, for example, that PA : PD equals the reciprocal of PB : PC,
or equals PC : PB, as shown above.
301. Secant to a Circle. A secant from an external j^oint to a
circle is understood to mean the segment of the secant that lies
between the given external point and the second j^oint of inter
section of the secant and circle.
178 BOOK III. PLANE GEOMETRY
Proposition XXII. Theorem
302. If from a point outside a circle a secant and a
tangent are draion, the tangent is the 7nean propor
tional hetiveen the secant and its external segment.
Given a tangent AD and a secant AC drawn from the point A to
the circle BCD.
To prove that AC. AD = AD: AB.
Proof. Draw DC and DB.
Now Z c is measured by ^ arc DB, § 214
and Z c' is measured by i arc DB. § 220
.'.Z.c = Z.c\
Then in the A ADC and ABD,
Z.a = /.a, Iden.
and Ac — Ac\
.. A^Z>C and ^5/)are similar. §286
.•.AC\AD=.AD'.AB, by §282. q.e.d.
303. Corollary. If from a fixed 'point outside a circle a
secant is dratvn, the product of the secant and its external
segment is constant in whatever direction the secant is drawn.
Smce AC:AD = AD:AB, §302
.. AC X AB = AD^ §261
Since AD is constant (§ 192), therefore AC x AB is constant.
NUMERICAL PROPERTIES OF FIGURES 179
Proposition XXIII. Theorem
304. The square on the bisector of an angle of a tri
angle is equal to the product of the sides of this angle
diminished hy the product of the segments made by the
bisector upon the third side of the triangle.
~C
D
Given the line CP bisecting the angle ACB of the triangle ABC.
To prove that CP'' = CA x BCAP x PB.
Proof. Circumscribe the OBCA about the A ABC. § 240
Produce CP to meet the circle in Z>, and draw BD.
Then in the A BCD and PC A,
and
/.m = Z. m\
Given
Aa' = A a.
§214
{Each is measured by I arc BC.)
•. the A BCD and PC A are similar.
§286
.. CD:CA=BC: CP.
§282
.. CAxBC=CDX CP
§261
= {CP\PD)CP
Ax. 9
= CP'' + CP X PD.
CPXPD=APXPB.
§299
.. CA xBC=CP^{APxPB.
Ax. 9
CP^ =CAxBC APx PB, by Ax. 2.
Q.E.D.
But
This theorem enables us to compute the bisectors of the angles of a
triangle terminated by the opposite sides, if the sides are known. The
theorem may be omitted without destroying the sequence.
180 BOOK III. PLANE GEOMETRY
Proposition XXIV. Theorem
305. In any triangle tlie jproduct of two sides is equal
to the product of the diameter of the circumscribed circle
hy the altitude upon the third side.
Given the triangle ABC with CP the altitude, ADBC the circle
circumscribed about the triangle ABC^ and CD a diameter.
To prove that CAxBC=CDx CP,
Proof. Draw BD.
Then in the A APC and DBC,
Z CPA is a rt. Z, Given
Z CBD is a rt. Z, § 215
Z <t is measured by \ arc BC,
and Z a' is measured by \ arc BC. § 214
.\Aa = Z.a\
.. A APC and D^C are similar. § 287
( Two rt. A are similar if an acute Z of the one is equal to an acute Z
of the other.)
.. CA : CD=CP:BC. § 282
.'. CA XBC=CDX CP, by § 261. q.e.d.
This theorem may be omitted without destroying the sequence. Props.
XXIII and XXIV are occasionally demanded in college entrance ex
aminations, but they are not necessary for proving subsequent propo
sitions or for any of the exercises. Teachers may therefore use their
judgment as to including them.
NUMERICAL PROPERTIES OF FIGURES 181
EXERCISE 47
1. The tangents to two intersecting circles, drawn from any
point in their common chord produced, are equal.
2. The common chord of two intersecting circles, if produced,
bisects their common tangents.
3. If two circles are tangent externally, the common internal
tangent bisects the two common external tangents.
4. If a line drawn from a vertex of a triangle divides the
opposite side into segments proportional to the adjacent sides,
the line bisects the angle at the vertex.
5. If three circles intersect one another, the common chords
are concurrent.
Let two of the chords, AB and CD, meet at 0. Join
the point of intersection E to 0, and suppose that EO
produced meets the same two circles at two different
points P and Q. Then prove that OP = OQ {^ 299),
and hence that the points P and Q coincide.
6. The square on the bisector of an exterior angle of a triangle
is equal to the product of the segments determined by this
bisector upon the opposite side, diminished by ^
the product of the other two sides. / '/^o^
Let CD bisect the exterior ZBCH of the A ABC. ^^
A ADC and FBC are similar (§ 286). Apply § 303.
7. If the line of centers of two circles meets the circles at the
consecutive points A, B, C, D, and meets the common external
tangent at P, then PA xPD = PBx PC.
8. The line of centers of two circles meets the common
external tangent at P, and a secant is drawn from P, cutting
the circles at the consecutive points E, F, G, II. Prove that
PE XPH=PFXPG.
Draw radii to the points of contact, and to E, P, G, H. Let fall Js on
PR from the centers of the ©. The various pairs of A are similar.
182 BOOK III. PLANE GEOMETEY
Proposition XXV. Problem
306. To divide a given line into parts proportional
to any number of given lines.
A M' n' B
^^ ^ V^N
\^^ \ \ \
\ "V \ \
n^\ \
"^ ^^\\
n p\
p Px
Given the lines AB^ m, n, and p.
Required to divide AB into parts proportional to m, n, and p.
Construction. Draw AX, making any convenient Z. with AB.
On AX take AM equal to m,
MN equal to n, and NP equal to j9.
Draw BP.
From N draw NN' II to PB,
and from M draw MM' II to PB.
Then M' and iV' are the division points required. q.e.f.
Proof. Through A draw a line II to PB.
AM' _ M'N' N'B
AM ~ MN ~ NP'
{Three or more  lines cut off proportional intercepts on any two
transversals.)
Substituting 7«,, n, and^^^ for their equals AM, MN, and NP,
AM' M'N' N'B , ^
we have = = Ax. 9
7)1 n p
This means that AB has been divided as required. q.e.d.
In like manner, we may divide AB into parts proportional to any
number of given lines.
PROBLEMS OF CONSTRUCTION 183
Proposition XXVI. Phoblem
307. To find the fourth proportional to three given
lines.
A____yii_ B n_ G__^
\r
Given the three lines m, n, and p.
Required to find the fourth proportional to /«, w, and p.
Construction. Draw two lines ^A' and AY containing any
convenient angle.
On AX take AB equal to m,
and take BC equal to n.
On ^ F take AD equal to p.
Draw BD.
From C draw CE II to BD, meeting A Y at E.
Then DE is the fourth proportional required. q.e.f.
Proof. AB : BC = AD : DE. § 273
(If a line is drawn through tioo sides o/a A II to the third side, it divides
the two sides proportionally.)
Substituting 7ti, n, and^ for their equals AB, BC, and AD,
we have m:n=2^ '• ^^ ^^ ^
Therefore DE is the fourth proportional to m, n, and p,
by § 259. Q.E.D.
308. Corollary. To find the third proportional to two
given lines.
In the above proof take m, w, n as the given lines instead of m, n, p.
184 BOOK III. PLANE GEOMETEY
Proposition XXVII. Problem:
309. To find the mean proportional between two given
lines.
/
H
\
\
1
m
!
n
!
!
A
m
n B
Given the two lines m and n.
Required to find the mean proportional between m and n.
Construction. Draw any line AE, and on AE take AC equal
to m, and CB equal to n.
On yli^ as a diameter describe a semicircle.
At C erect the _L CH, meeting the circle at //.
Then CH is the mean proportional between m and n. q.e.f.
Proof. AC :CH=CH: CB. § 297
( The A. from any point on a circle to a diameter is the mean
proportional between the segments of the diameter.)
Substituting for AC and CB their equals m and n,
we have m : CH= CH : n, by Ax. 9. q.e.d.
310. Extreme and Mean Ratio. If a line is divided into two
segments such that one segment is the mean proportional be
tween the whole line and the other segment, the line is said to
be divided in extreme and mean ratio.
E.g. the line a is divided in extreme and mean ratio, if a segment x
is found such that
a : X = X : a — X.
The division of a line in extreme and mean ratio is often called the
Golden Section.
PROBLEMS OF CONSTRUCTION 185
Proposition XXVIII. Problem
311. To divide a given line in extreme and mean ratio.
:>Q
^1
,.""nv
A C B
Given the line AB.
Recfiired to divide AB in extreme and mean ratio.
Construction. At B erect a ± BE equal to half of AB.
From £^ as a center, with a radius equal to EB, describe a O.
Draw AE, meeting the circle at F and G.
On ^^ take ^C equal to .IF.
On BA produced take A C ' equal to A G.
Then AB \^ divided internally at C and externally at C in
extreme and mean ratio.
That is, AB : AC = AC : CB, and AB : AC = AC : C'B. q.e.f.
Proof. AG:AB = A B : A F. § 302
From AG:AB = AB:AF,
AGAB:AB =
ABAF:AF.
.'.AGFG:AB =
ABAC:AC.
.'. AC:AB = CB:AC.
.•.AB:AC = AC:CB,
by inversion, § 266. q.e.d.
.'. AB:AG = AF: AB. § 266
.'. AB\AG: AG =
AF\AB: AB.
.\ABAAC:AC =
AF + FG'.AB.
.'. CB:AC = AC: AB.
.. AB:AC = AC:C'B,
by §§261, 264. q.e.d.
186 BOOK III. PLANE GEOMETRY
Proposition XXIX. Pkoblem
312. Upon a given line correspondincj to a given side
of a given polygon, to construct a polygon similar to
the given polygon.
Y
I
I
z /i \
\ / I \ ,^
\ / " /
Given the line A^B^ and the polygon ABCDE.
Required to construct on A'B\ corresponding to AB, a
polygon similar to the polygon ABCDE.
Construction. From A draw the diagonals AD and AC.
From A ' draw A 'X, A'Y, and A 'Z, making Ax', y\ and z' equal
respectively to A x, y, and «. § 232
From B' draw a line, making Z B' equal to Z B,
and meeting A'X at C'.
From C" draw a line, making Z D'C'B' equal to Z DCB,
and meeting A'Y sX D'.
From i)' draw a line, making ZE'D'C equal to ZEDC,
and meeting ^'Z at E'.
Then A'B'C'D'E' is the required polygon. qe.f.
Proof. The A ABC and ^'^'C, the AACD and yl'C'/)', and
the A ADE and ^'Z>'J5:', are similar. § 286
Therefore the two polygons are similar, by § 293. q.e.d.
EXERCISES 187
EXERCISE 48
1. If a and h are two given lines, construct a line equal to ar,
where x = 'Wab. Consider the special case of « = 2, ^ = 3.
2. If m and n are two given lines, construct a line equal to x,
where x = V 2 mn.
3. Determine both by geometric construction and arith
metically the third proportional to the lines 1^ in. and 2 in.
4. Determine both by geometric construction and arith
metically the third proportional to the lines 4 in. and 3 in.
5. Determine both by geometric construction and arith^
metically the fourth proportional to the lines li in., 2 in.',
and 21 in.
6. Determine both by geometric construction and arith
metically the mean proportional between the lines 1.2 in.
and 2.7 in.
7. Eind geometrically the square root of 5. Measure the
liiie and thus determine the approximate arithmetical value.
8. A map is drawn to the scale of 1 in. to 50 mi. How far
apart are two places that are 2^\ in. apart on the map ?
9. Eind by geometric construction and arithmetically the
third proportional to the two lines ly^^ in. and 2 in.
10. Divide a line 1 in. long in extreme and mean ratio.
Measure the two segments and determine their lengths to
the nearest sixteenth of an inch.
11. Divide a line 5 in. long in extreme and mean ratio.
Measure the two segments and determine their lengths to
the nearest sixteenth of an inch.
12. Divide a line 6 in. long in extreme and mean ratio.
Measure the two segments and determine their lengths to
the nearest sixteenth of an inch.
The propositions on this page are taken from recent college entrance
examination papers.
188
BOOK III. PLANE GEOMETRY
13. Through a given point P within a given circle to draw
a chord AB ^o that the ratio AP : BP shall equal
a given ratio w : n.
Draw OPC so that OP.PC = n: m.
Draw CA equal to the fourth proportional to n, w,
and the radius of the circle.
14. To draw two lines making an angle of 60°, and to con
struct all the circles of ^ in. radius that are tangent to both lines.
15. To draw through a given point P in the
arc subtended by a chord ABsl chord which shall
be bisected by AB. e^
On radius OP take CD equal to CP. Draw DE II to BA.
16. To construct two circles of radii ^ in. and 1 in. respec
tively, which shall be tangent externally, and to construct a
third circle of radius 3 in., which shall be tangent to each of
these two circles and inclose both of them.
17. To draw through a given external
point P a secant PAB to a given circle so
that the ratio PA : AB shall equal the given
ratio 7)1 : n.
Draw the tangent PC. Make PD . DC = m . n. PA : PC = PC : PB.
18. To draw through a given external point
P a secant PAB to a given circle so that
AB^ = PA X PB.
19. An equilateral triangle ABC is 2 in.
on a side. To construct a circle which shall be
tangent to A B at the point A and shall pass through the point C.
20. To draw through one of the points
of intersection of two circles a secant so
that the two chords that are formed shall
be in the given ratio m : n.
P^~,
EXERCISES 189
21. In a circle of 3 in. radius chords are drawn through a
point 1 in. from the center. What is the product of the seg
ments of these chords ?
22. The chord AB is S in. long, and it is produced through
B to the point P so that PB is equal to 12 in. Find the tangent
from P.
23. Two lines AB and CD intersect at 0. How would you
ascertain, by measuring OA, OB, OC, and OD, whether or not
the four points A, B, C, and D lie on the same circle ?
24. This figure represents an instrument for finding the
centers of circular plates or sections of shafts. OC is a ruler
that bisects the angle A OB, and A
and OB are equal. Show that, if A
and B rest on the circle, OC passes
through the center, and that by
drawing two lines the center can be
found.
25. If three circles are tangent externally each to the other
two, the tangents at their points of contact pass through the
center of the circle inscribed in the triangle formed by joining
the centers of the three given circles.
26. In the isosceles triangle ABC, C is a right angle, and ^C
is 4 in. With A as center and a radius 2 in. a circle is described.
Required to describe another circle tangent to the first and also
tangent to BC at the point B.
27. Find the center of a circle of ^ in. radius, so drawn in
a semicircle of radius 2 in. as to be tangent to the semi
circle itself and to its diameter
28. To inscribe in a given circle a triangle similar to a given
triangle. , ,  r .c
29. To draw two straight linesegments, having given their
sum and their ratio. .... .. ,
190 BOOK III. PLANE GEOMETRY
EXERCISE 49
Review Questions
1. What is meant by ratio ? by proportion ?
2. If a'.b=^c: d, write four other proportions involving
these quantities.
3. If a : b = c : d, is it true in general that a{l:b \l
z=c:d? Is it ever true ?
4. When is a line divided harmonically ? The bisectors of
what angles of a triangle divide the opposite side harmonically ?
5. What are the two conditions necessary for the similarity
of two polygons ?
6. Are two mutually equiangular triangles similar ? Are
two mutually equiangular polygons always similar ?
7. Are two triangles similar if their corresponding sides
are proportional ? Are two polygons always similar if their
corresponding sides are proportional ?
8. If two triangles have their sides respectively parallel,
are they similar ? Is this true of polygons in general ?
9. If two triangles have their sides respectively perpendicu
lar, are they similar ? Is this true of polygons m general ?
10. Complete in two ways : The perimeters of two similar
polygons have the same ratio as any two corresponding ••• .
11. If in a right triangle a perpendicular is drawn from
the vertex of the right angle to the hypotenuse, state three
geometric truths that follow.
12. If two secants intersect outside, on, or within a circle,
what geometric truth follows ?
"13. How would you proceed to divide a straight line into
seven equal parts ?
 14. How. would you proceed to find the square root of 7 by
measuring the length of a line ? ' ;. ■
BOOK IV
AREAS OF POLYGONS
313. Unit of Surface. A square the side of which is a unit of
length is called a unit of surface.
Thus a square that is 1 inch long is 1 square inch, and a square that is
1 mile long is 1 square mile. If we are measuring the dimensions of a
room in feet, we measure the surface of the floor in square feet. In
the same way we may measure the page of this book in square inches
and the area of a state in square miles.
314. Area of a Surface. The measure of a surface, expressed
in units of surface, is called its area.
If a room is 20 feet long and 15 feet wide, the floor contains 300 square
feet. Therefore the area of the floor is 300 square feet. Usually the two
sides of a rectangle are not commensurable, although by means of frac
tions we may measure them to any required degree of approximation.
The incommensurable cases in theorems like Prop. I of this Book may
be omitted without interfering with the sequence of the course.
315. Equivalent Figures. Plane figures that have equal areas
are said to be equivalent.
In propositions relating to areas the words rectangle^ triangle^ etc., are
often used for area of rectangle^ area of triangle^ etc.
Since congruent figures may be made to coincide, congruent figures
are manifestly equivalent.
Because their areas are equal, equivalent figures are frequently spoken
of as equal figures. The symbol = is used both for " equivalent " and for
" congruent," the sense determining which meaning is to be assigned to it.
Occasionally these symbols are used : = , = , or = for congruent, = for
equal, and =o for equivalent.
Since the word congruent means " identically equal," the word equal is
often used to mean ''equivalent."
191
192 BOOK IV. PLANE GEOMETRY
Proposition I. Theorem
316. Two rectangles having equal altitudes are to each
other as their bases.
D
x B A X
Given the rectangles AC and AF^ having equal altitudes AD.
To prove that \I}AC:\I\AF = base AB : base AE.
Case 1. When AB and AE are commensurable.
Proof. Suppose AB and AE have a common measure, as AX.
Suppose AX '\?> contained m times in AB and n times in AE.
Then AB:AE = m:n.
(For m and n are the numerical measures of AB and AE.)
Apply ^X as a unit of measure to ^^ and AE, and at the
several points of division erect Js.
These Js are all ± to the upper bases, § 97
and these Js are all equal. § 128
Since to each base equal to AX there is one rectangle,
.'. □ .4C is divided into m rectangles,
and □ .4F is divided into n rectangles. § 119
These rectangles are all congruent. § 133
., ]Z\AC'.L^AF = m:n.
.'. {3 AC '.n^AF^^AB'.AE, by Ax. 8. q.e.d.
In this proposition we again meet the incommensurable case, as on
pages 116 and 157. This case is considered on page 193 and may be
omitted without destroying the sequence of the propositions.
AREAS OF POLYGONS
193
Case 2. When AB and AE are incommensurable.
D
D
Proof. Divide A E into any number of equal parts, and apply
one of these parts to AB as many times as AB will contain it.
Since AB and AE are incommensurable, a certain number of
these parts will extend from A to some point P, leaving a re
mainder PB less than one of them. Draw PQ _L to AB.
□ .4Q AP
\Z\AF~ AE'
Then
Case 1
By increasing the number of equal parts into which AE is
divided we can diminish the length of each, and therefore can
make PB less than any assigned positive value, however small.
Hence PB approaches zero as a limit, as the number of parts
of AE is indefinitely increased, and at the same time the cor
responding □ PC approaches zero as a limit. § 204
Therefore AP approaches yl^ as a limit, and D^Q ap
proaches \Z\ AC as a limit.
.*. the variable — — approaches — — as a limit,
ACt A. E
and the variable 7=— — approaches 7=— — as a limit.
A P CH A Q
But — — is always equal to ^^\ „ ? as AP varies in value and
\3AF
approaches AB as a limit.
n\AC AB ^ , ^_
Case 1
Q.E.D.
317. Corollary. Two rectangles having equal bases are to
each other as their altitudes.
194 BOOK IV. PLANE GEOMETRY
Proposition II. Theorem
318. Tivo rectangles are to each other as the j^roducts
of their bases hy their altitudes.
R
b U b
Given the rectangles i? and R\ having for the numerical measure
of their bases b and h\ and of their altitudes a and a! respectively.
To prove that — = — •
R a'b'
Proof. Construct the rectangle S, with its base equal to that
of R, and its altitude equal to that of R\
R_a
' R'~b''
Then
and
§317
§316
Since we are considering areas, we may treat R, S, and S' as
numbers and take the products of the corresponding members
of these equations. ,, ^ § 272
We therefore have
n_
R'
ab
by Ax. 3.
Q.E.D.
319. Products of Lines. When we speak of the product of a
and b we mean the product of their numerical
values. It is possible, however, to think of a
line as the product of two lines, by changing
the definition of multiplication. Thus in this
fig\ire in which two parallels are cut by two intersecting trans
versals, we have l:a = b : x. Therefore x — ah. In the same
way we may find xc^ or abc^ the product of three lines.
AREAS OF POLYGONS
Proposition III. Theorem
195
320. The area of a rectangle is equal to the product
of its base hy its altitude.
Q
Given the rectangle /?, having for the numerical measure of its
base and altitude b and a respectively.
To prove that the area of R — ah.
Proof. Let U be the unit of surface. § 313
Then
ab
1x1
ah.
§318
R
But J = the number of units of surface in R, i.e. the area
oiR. §314
.". the area of R = ab, by Ax. 8. q.e.d.
321. Practical Measures. When the base and altitude both
contain the linear unit an integral number of times, this propo
sition is rendered evident by dividing the rectangle into squares,
each equal to the unit of surface.
Thus, if the base contains seven hnear
units and the altitude four, the rectangle
may be divided into twentyeight squares,
each equal to the unit of surface. Practi
cally this is the way in which we conceive
the measure of all rectangles. Even if the
sides are incommensurable, we cannot determine this by any measuring
instrument. If they seem to be incommensurable with a unit of a
thousandth of an inch, they might not seem to be incommensurable with
a unit of a millionth of an inch.
196 BOOK IV. PLANE GEOMETRY
EXERCISE 50
1. A square and a rectangle have equal perimeters, 144
yd., and the length of the rectangle is five times the breadth.
Compare the areas of the square and rectangle.
2. On a certain map the linear scale is 1 in. to 10 mi.
How many acres are represented by a square  in. on a side ?
3. Find the ratio of a lot 90 ft. long by 60 ft. wide to a
field 40 rd. long by 20 rd. wide.
4. Find the area of a gravel walk 3 ft. 6 in. wide, which
surrounds a rectangular plot of grass 40 ft. long and 25 ft.
wide. Make a drawing to scale before beginning to compute.
5. Find the number of square inches in this
cross section of an L beam, the thickness being J in. '
6. What is the perimeter of a square field that
contains exactly an acre ? '>^2%^
7. A machine for planing iron plates planes a space ^ in.
wide and 18 ft. long in 1 min. How long will it take to plane
a plate 22 ft. 6 in. long and 4 ft. 6 in. wide, allowing 51 min.
for adjusting the machine ?
8. How many tiles, each 8 in. square, will it take to cover
a floor 24 ft. 8 in. long by 16 ft. wide ?
9. A rectangle having an area of 48 sq. in. is three times
as long as wide. What are the dimensions ?
10. The length of a rectangle is four times the width. If
the perimeter is 60 ft., what is the area?
11. From two adjacent sides of a rectangular field 60 rd.
long and 40 rd. wide a road is cut 4 rd. wide. How many
acres are cut off for the road ?
12. From one end of a rectangular sheet of iron 10 in. long
a square piece is cut off leaving 25 sq. in. in the rest of the
sheet. How wide is the sheet ?
AREAS OF POLYGONS
Proposition IV. Theorem
197
322. Tlie area of a parallelogram is equal to the
product of its base by its altitude.
YD X G
A b B A b B
Given the parallelogram ABCD^ with base h and altitude a.
To prove that the area of the CJABCD = ah.
Proof. From B draw BX 1. to CD or to CD produced, and
from A draw ^ F J_ to CD produced.
Then ABXY is a rectangle, with base b and altitude a.
Since AY= BX, 2indAD = BC, § 125
..the rt. A ADY and BCX are congruent. § 89
From ABC Y take the A BCX; the \Z1ABXY is left.
From ABC Y take the A ADY; the EJABCD is left.
.. n\ABXY= CJABCD. Ax. 2
But the area of the □ A BX Y = ah. § 320
.*. the area of the CJABCD = ah, by Ax. 8. q.e.d.
323. C'oKOLLARY 1. P aralUlograms having equal bases and
equal altitudes are equivalent.
324. Corollary 2. Parallelograms having equal bases are
to each other as their altitudes; parallelograms having equal
altitudes are to each other as their bases ; any two jmrallelo
grams are to each other as the products of their bases by
their altitudes.
This was regarded as veiy interesting by the ancients, since an ignorant
person might think it impossible that the areas of two parallelograms
could remain the same although their perimeters differed without limit.
198 BOOK IV. PLANE GEOMETRY
Proposition V. Theorem
325. The area of a triangle is equal to half the
product of its base by its altitude.
A b B
Given the triangle ABC, with altitude a and base b.
To prove that the area of the A ABC = ^ab.
Proof. With AB and BC SiS adjacent sides construct the
parallelogram A BCD. § 238
Then A ABC = i EJABCD. § 126
But the area of the EJABCD = ah. § 322
.*. the area of the A ABC = \ ab, by Ax. 4. q.e.d.
326. Corollary 1. Triangles having equal bases and equal
altitudes are equivalent.
327. Corollary 2. Triangles having equal bases are to each
other as their altitudes ; triangles having equal altitudes are to
each other as their bases ; any two triangles are to each other
as the products of their bases by their altitudes.
Has this been proved for rectangles ? What is the relation of a triangle
to a rectangle of equal base and equal altitude ? What must then be the
relations of triangles to one another ?
328. Corollary 3. The product of the sides of a right tri
angle is equal to the product of the hypotenuse by the altitude
from the vertex of the right angle.
How is the area of a right triangle found in terms of the sides of the
right angle ? in terms of the hypotenuse and altitude ? How do these
results compare ?
AREAS OF POLYGONS
199
Proposition VI. Theorem
329. The area of a trapezoid is equal to half the
product of the sum of its bases by its altitude.
A b B
Given the trapezoid ABCD^ with bases b and V and altitude a.
To prove that the area of ABCD ='^a(h\ h'}.
Proof. Diaw the diagonal A C.
Then the area of the AABC = ^ ah,
and the area of the A A CZ) = ^ ah'. § 325
.*. the area of ABCD = ^ a (^ + h'), by Ax. 1. q.e.d.
330. Corollary. The area of a trapezoid is equal to the
product of the line joining the midpoints of its nonparallel
sides hy its altitude.
How is the line joining the midpoints of the nonparallel sides related
to the sum of the bases (§ 137) ?
331. Area of an Irregular Polygon. The area of an irregular
polygon may be found by dividing the polygon into triangles,
and then finding the area of each
of these triangles separately.
A common method used in land sur
veying is as follows : Draw the longest
diagonal, and let fall perpendiculars
upon this diagonal from the other ver
tices of the polygon. The sum of the
right triangles, rectahglep, and trapezoids is equivalent to the polygon.
200 BOOK IV. PLANE GEOMETEY
EXERCISE 51
Find the areas of the parallelograms whose bases and
altitudes are respectively as follows :
1. 2.25 in., 1 J in. 3. 2.7 ft., 1.2 ft. 5. 2 ft. 3 in., 7 in.
2. 3.44 in., li in. 4. 5.6 ft., 2.3 ft. 6. 3 ft. 6 in., 2 ft.
Find the areas of the triangles whose bases and altitudes ai'e
respectively as follows :
7. 1.4 in., 11 in. 9. 6^ ft., 3 ft. 11. 1 ft. 6 in., 8 in.
8. 2.5 in., 0.8 in. 10. 5.4 ft., 1.2 ft. 12. 3 ft. 8 in., 3 ft.
Find the areas of the trapezoids ivhose bases are the first
two of the following numbers, and whose altitudes are the
third numbers :
13. 2 ft., 1 ft., 6 in. 15. 3 ft. 7 in., 2 ft., 14 in.
14. 2^ ft., 11 ft., 9 in. 16. 5 ft. 6 in., 3 ft., 2 ft.
Find the altitudes of the parallelograms whose areas and
bases are respectively as follows :
17. 10 sq. in., 5 in. 19. 28 sq. ft., 7 ft. 21. 30 sq. ft., 12 ft.
18. 6sq. in., 6in. 20. 27 sq. ft., 6 ft. 22. 80 sq. in., 16 in.
Find the altitudes of the triangles whose areas and bases
are respectively as follows :
23. 49 sq. in., 14 in. 25. 50 sq. ft., 10 ft. 27. 110 sq. yd., 10 yd.
24. 48 sq. in., 12 in. 26. 160 sq. ft., 20 ft. 28. 176 sq. yd., 32 yd.
Find the altitudes of the trapezoids whose areas and bases
are respectively as follows :
29. 33 sq. in., 5 in., 6 in. 31. 13 sq. ft., 9 ft., 5 ft.
30. 15 sq. in., 4 in., 6 in. 32. 70 sq. yd., 9 yd., 11 yd. ^. ,
AREAS OF POLYGONS 201
Proposition YII. Theorem
332. The areas of two triangles that have an angle
of the one equal to an angle of the other are to each
other as the 2^roducts of the sides including the equal
angles.
Given the triangles ABC and ADE^ with the common angle A,
^ , , A ABC ABxAC
To prove that 7— rrrr. = ~7T^ 7~r,
^ AADE ADxAE
Proof. Draw BE.
Then
A ABC AC
A ABE AE
A ABE AB ^ ^^„
{Triangleff having equal altitudes are to each other as their bases.)
Since we are considering numerical measures of area and
length, we may treat all of the terms of these proportions as,
numbers.
Taking the product of the first members and the product of
the second members of these equations, we have
AABE X A ABC AB x AC
A ADE X A ABE AD X AE
That is, by canceling AABE, we have the proportion
Ax. 3
AABC ABxAC
= O.E.D
AADE ADXAE
202 BOOK IV. PLANE GEOMETRY
Proposition VIII. Theorem
333. The areas of tiuo similar triangles are to each
other as the squares on any two corresjjonding sides.
A B A'
Given the similar triangles ABC and A'B'C\
^ ^, ^ A ABC AB'
To prove that .fp,^> = ■ 2 •
Proof. Since the triangles are similar, Given
.\AA=AA'. §282
_, A ABC ABxAC
^^^^ aJ^B^'^A'B'xA'C' ^^^^
( The areas of two triangles that have an angle of the one equal to
an angle of the other are to each other as the products
of the sides including the equal angles.)
. AABC _ AB_ AC_
Ihatis, AA'B'C' A'B'^A'C'
But ^ = ~ • §282
A'B' A'C
{Similar polygons have their corresponding sides proportional.)
Substituting — — , for its equal 7777,' we have
A J:> A Ly
AABC _ AB_ AB_
AA'B'C' A'B' ^ A'B'' ^^
AABC AB^
AREAS OF POLYGONS 203
Proposition IX. Theorem
334. The areas of two similar polygons are to each
other as the squares on any tioo corresponding sides.
Given the similar polygons ABODE and A'B^C'D'E\ of area 5
and s' respectively.
To prove that
s:s' = AB'':A'B'\
Proof. By drawing all the diagonals from any correspond
ing vertices .4 and A', the two similar polygons are divided
into similar triangles.
AADE ad' AACB AC'' A ABC
AA'D'E' Ji^r^ AA'C'D' jJc'
AADE AACD
That is,
AA'D'E' AA'C'D'
AADE{AACD\AABC
•' AA'B'C
_ A ABC
~ AA'B'C'
A ABC
AA'D'E' + A A'C'D' + A A'B'C AA'B'C
:2 TT^2
§292
ab'
A'B''
• §333
Ax. 8
ab'
§269
s:s' = AB :A'B' , by Ax. 11.
Q.E.D.
335. Corollary 1. The areas of tivo similar polygons are
to each other as the squares on any two corresponding lines.
336. Corollary 2. Corresponding sides of hvo similar poly
gons have the same ratio as the square roots of the areas.
204 BOOK IV. PLANE GEOMETKY
Proposition X. Theorem
337. Tlie square on the hypotenuse of a right triangle
is equivalent to the sum of the squares on the other tivo
sides.
/
R
X S
Given the right triangle ABC^ with AS the square on the hypote
nuse, and BN^ CQ the squares on the other two sides.
To prove that AS=BN\CQ.
Proof. Draw CX through C II to ^.S". § 233
Draw CR and BQ.
Since A c and x are rt. A, the Z PCB is a straight angle, § 34
and the line PCB is a straight line.
Similarly, the line A CN is a straight line.
In the A ARC and ABQ,
AR = AB,
AC = AQ,
and ARAC = ZBAQ.
{Each is the sum of a rt. Z and the Z BAC.)
.'. A ARC is congruent to A ABQ.
Furthermore the □ AX is double the A ARC.
{They have the same base AR, and the same altitude RX.)
§43
^65
Ax. 1
§68
§325
KUMERICAL PROPERTIES OF FIGURES 205
Again the square CQ is double the AABQ. § 325
{They have the same base AQ, and the same altitude AC.)
.'. the \I2AX is equivalent to the square CQ. Ax. 3
In like manner, by drawing CS and AM, it may be proved
that the rectangle BX is equivalent to the square BN.
Since square AS = nBX^[D AX, Ax. 11
.. AS = BN{ CQ, by Ax. 9. q.e.d.
The first proof of this theorem is usually attributed to Pythagoras (about
525 B.C.), although the truth of the proposition was known earlier. It is
one of the most important propositions of geometry. Various proofs may
be given, but the on6 here used is the most common. This proof is attrib
uted to Euclid (about 300 b.c), a famous Greek geometer.
338. Corollary 1. The square on either side of a right
triangle is equivalent to the difference of the square on the
hypotemise and the square on the other side.
339. Corollary 2. The diagonal and a side of a square
are incommensurable. jy ^
For AC^ = AB^ + BC^ = 2 AB^.
.: AC = ABV2.
Since V2 may be carried to as many decimal places as
we please, but cannot be exactly expressed as a rational
fraction, it has no common measure with 1. That is, AC :AB = V2, an
incommensurable number.
340. Projection. If from the extremities of a linesegment
perpendiculars are let fall upon another line, the segment thus
cut off is called the projection of the first line upon the second.
Thus C^iy is the projection of CD upon AB, or V is the projection
of I upon AB.
In general it is convenient to designate
by the small letter a the side of a triangle
opposite ZA, and so for the other sides; to
designate the projection of a by a'; and to
designate the height (altitude) by A. A c^ d'
206 BOOK IV. PLANE GEOMETRY
EXERCISE 52
Given the sides of a right triangle as follotvs^ find the
hypotenuse to two decimal places:
1. 30 ft., 40 ft. 3. 20 ft., 30 ft. 5. 2 ft. 6 in., 3 ft.
2. 45 ft., 60 ft. 4. 1.5 in., 2.5 in. 6. 3 ft. 8 in., 2 ft.
Given the hypotenuse and one side of a right triangle as
follows, find the other side to two decimal places :
7. 50 ft, 40 ft. 9. 10 ft., 6 ft. 11. 3 ft. 4 in., 2 ft.
8. 35 ft., 21 ft. 10. 1.2 in., 0.8 in. 12. 6 ft. 2 in., 5 ft.
13. A ladder 38 ft. 6 in. long is placed against a wall, with
its foot 23.1 ft. from the base of the wall. How high does it
reach on the wall ?
14. Find the altitude of an equilateral triangle with side s.
15. Find the side of an equilateral triangle with altitude h.
16. The area of an equilateral triangle with side s is i s^ V3.
17. Find the length of the longest chord and of the shortest
chord that can be drawn through a point 1 ft. from the center
of a circle whose radius is 20 in.
18. The radius of a circle is 5 in. Through a point 3 in.
from the center a diameter is drawn, and also a chord perpen
dicular to the diameter. Find the length of this chord, and
the distance (to two decimal places) from one end of the chord
to the ends of the diameter.
19. In this figure the angle C is a >/l\
right angle. From the relations AC = y^ \ \
AB X AF (§ 294) ^d C^= AB x BF, ^^ X I \ ^
show that lc'+ c]b% Zb^ ^
20. If the diagonals of a quadrilateral intersect at right
angles, the sum of the squares on one pair of opposite sides
is equivalent to the sum of the squares on the other pair.
NUMERICAL PROPERTIES OF FIGURES 207
Proposition XI. Theorem
341. In any triangle the square on the side opposite
an acute angle is equivalent to the sum of the squares
on the other two sides diminished hy ttvice the product
of one of those sides hy the projection of the other upon
that side.
c
Fig. 1
Given the triangle ABC^ A being an acute angle, and a' and V
being the projections of a and b resi)ectively upon c.
To prove that d^ = ¥\c^—2 h'c.
Proof. If D, the foot of the _L from C, falls upon c (Fig. 1),
a' = cb'.
If D falls upon c produced (Fig. 2),
a' = b'c.
In either case, by squaring, we have
a'2 = Z»'2_f_^2_2^'c. Ax. 5
Adding A^ to each side of this equation, we have
h^^a'^ = h^{b'^ + c^2b'c. Ax. 1
But h^ + «'2 = a^, and h^ + h"" = b\ § 337
Putting a^ and b'^ for their equals in the above equation, we
have
a^ = b^^c^2b'c, by Ax. 9.
Q.E.D.
208 BOOK IV. PLANE GEOMETRY
Proposition XII. Theorem
342. In any obtuse triangle the square on the side
opposite the obtuse angle is equivalent to the sum of
the squares on the other tivo sides increased by tivice
the product of one of those sides by the projection
of the other upon that side.
Given the obtuse triangle ABC^ A being the obtuse angle, and a'
and b' the projections of a and h respectively upon c.
To prove that a^ = b^ + e'\2 b'c.
Proof. a' = b' \c.
Squaring, a''' = b'"" + e" { 2 b'c.
Adding h^ to each side of this equation, we have
7,2 f ,,'2 =h^^ z,'2 4_ ^.2 _^ 2 b'c.
But h^ + a'^ = a^, and Ii^ + b'^ = b\
Ax. 11
Ax. 5
Ax. 1
§337
Putting (i^^iYidi V^ for their equals in the above equation, we have
a^ = b^^^^2b'c,\^j Ky.. 9.
Q.E.D.
Discussion. By the Principle of Continuity the last three theorems
may be included in one theorem by letting the A A change from an
acute angle to a right angle and then to an obtuse angle. Let the
student explain.
The last three theorems enable us to compute the altitudes of a tri
angle if the three sides are known ; for in Prop. XII we can find 6', and
from h and V we can find h.
NUMERICAL PROPERTIES OF FIGURES 209
EXERCISE 53
Find the lengths^ to two decimal places, of the diagonals of
the squares whose sides are :
1. 7 in. 2. 10 in. 3. 9.2 in. 4. 1 ft. 6 in. 5. 2 ft. 3 in.
Find the lengths, to two decimal places, of the sides of
the squares whose diagonals are :
6. 4 in. 7. 8 in. 8. 5 ft. 9. VE in. 10. 2 ft. 6 in.
11. The minute hand and hour hand of a clock are 6 in. and
4^ in. long respectively. How far apart are the ends of the
hands at 9 o'clock ?
12. A rectangle whose base is 9 and diagonal 15 has the
same area as a square whose side is x. Find the value of x.
13. A ring is screwed into a ceiling in a room 10 ft. high.
Two rings are screwed into the floor at points 5 ft. and 12 ft.
from a point directly beneath the one in the ceiling. Wires
are stretched from the ceiling ring to each floor ring. How
long are the wires ? (Answer to two decimal places.)
14. The sum of the squares on the segments of two perpen
dicular chords is equivalent to the square on the diameter of
the circle.
If AB, CD are the chords, draw the diameter BE, and draw AC,
ED, BD. Prove that AC = ED.
15. The difference of the squares on two sides of a triangle
is equivalent to the difference of the squares on the segments
of the third side, made by the perpendicular on the third
side from the opposite vertex.
16. In an isosceles triangle the square on one of the equal
sides is equivalent to the square on any line drawn from the
vertex to the base, increased by the product of the segments
of the base.
210
BOOK IV. PLANE GEOMETRY
Proposition XIII. Theorem
343. The sum of the squares on tioo sides of a tri
angle is equivalent to tivice the square o?i half the third
side, increased hy tivice the square on the median
upon that side.
The difference of the squares on tioo sides of a tri
angle is equivalent to tivice the j^^oduct of the third
side hy the projection of the median upon that side.
Given the triangle ABC^ the median m, and mJ the projection of
m upon the side a. Also let c be greater than b.
To prove that 1. c' + 6' = 2 .RE' + 2 w%
2. c'l>' = 2am'.
■ Proof. The Z AMB is obtuse, and the Z CMA is acute. § 116
Since c>b, M lies between B and D. § 84
Then
and
c^ = BM ' + m^ \2BMm',
IP' = MC^ \m^2MC ' m'.
§342
§341
Adding these equals, and observing that BiM = MC, we have
c'i_^p = 2 BM^ + 2 m^. Ax. 1
Subtracting the second from the first, we have
c^ — b^ = 2 am', by Ax. 2. q.e.d.
Discussion. Consider the proposition when c = h.
This theorem may be omitted without interfering with the regular
sequence. It enables us to compute the medians when the three sides
are known.
EXERCISES 211
EXERCISE 54
1. To compute the area of a triangle in terms of its sides.
c
A C B D
At least one of the angles ^ or B is acute. Suppose A is acute.
In the A A DC, h^ = b"AD^. Why ?
In the A ABC, a^ = ¥ { c^  2c x AD. Why ?
62 4 c2  a2
Therefore AD
Hence h^ = b
2c
(62 4. c2 _ a2)2 4 ^2^2 _ (52 4. c2 _ a2)2
4c2 4c2
_ (2 6c + 62 ^ (.2  a2) (2 6c  62 _ c2 + a2)
" ~ 4c2
_ {(6 + c)2  a2} {a2  (6  c)2}
~ 4c2
_ (g + 6 + c) (6 + c — g) (g + 6 — c) (g — 6 + c)
 4c2"'
Let g+6 + c = 2 5, where 5 stands for semiperimeter.
Then 6 + c — g = g + 6 + c — 2g = 2s — 2g = 2(s — g).
Similarly • g + 6 — c = 2 (s — c),
and g6 + c = 2(s6).
^„ 2 s X 2 (.s  g) X 2 (8  6) X 2 (s  c)
Hence h^ = ^^ ^^ ^^ •
4c2
By simplifying, and extracting the square root,
2 /
hz= Vs{s — a) (s — 6) (s — c).
Hence the area = lch= Vs [s — a) {s — 6) {s — c).
For example, if the sides arc 3, 4, and 6,
area = V6(6  3) (6  4) (6  5) = V6 • 3 • 2 = 6.
212 BOOK IV. PLANE GEOMETRY
If Ex. 1 has been studied, find the areas, to two decimal
places, of the triangles whose sides are :
2. 4, 5, 6. 4. 6, 8, 10. 6. 7, 8, 11. 8. 1.2, 3, 2.1.
3. 5, 6, 7. 5. 6, 8, 9. 7. 9, 10, 11. 9. 11, 12, 13.
10. To compute the radius of the circle circumscribed about
a triangle in terms of the sides of the triangle. (Solve only if
§ 305 and Ex. 1 have been taken.)
Let CD be a diameter.
By § 305, what do we know about the products CA x BC and CD x CP ?
What does this tell us of ab and 2 r • CP, r be
ing the radius ? X^ ^^^^^^
From Ex. 1, what does CP equal in terms of / ^^^^^^/ i ^
the sides ? ^[ ^ c — 7^ — h~^^
Is it therefore possible to show that I / ^^^j
abc „ \ /'"^^ J
4 Vs (s — a) (8 — 6) (s — c) .dV_ ^^
If Exs. 1 and 10 have been studied, compute the radii, to
two decimal places, of the circles circumscribed about the
triangles whose sides are :
11. 3, 4, 5. 12. 27, 36, 45. 13. 7, 9, 11. 14. 10, 11, 12.
15. To compute the medians of a triangle in terms of its sides.
Omit if § 343 has not been taken. What do we know
about a^ + 6^ as compared with 2 m^ j 2 (  J ?
From this relation show that
If Ex. 15 has been studied, compute the three medians, to
two decimal places, of the triangles whose sides are :
16. 3, 4, 5. 17. 6, 8, 10. 18. 6, 7, 8. 19. 7, 9, 11.
20. If the sides of a triangle are 7, 9, and 11, is the angle
opposite the side 11 right, acute, or obtuse ?
EXERCISES
213
1^
I
I .
H
.H
\C
21. The square constructed upon the sum of two lines is
equivalent to the sum of the squares constructed upon these
two lines, increased by twice the rectangle of these lines.
Given the two lines AB and BC, and AC their sum. Construct the
squares AKGC and ADEB upon AC and AB respec
tively. Produce BE and DE to meet KG and CG in H ^ B c
and F respectively. Then we have the square EHGF,
with sides each equal to BC. Hence the square AKGC
is the sum of the squares ADEB and EHGF, and the
rectangles DKHE and BEFC.
This proves geometrically the algebraic formula
(a + 6)2 = a2 + 2 a6 + b^.
22. The square constructed upon the difference of two lines
is equivalent to the sum of the squares constructed upon these
two lines, diminished by twice their rectangle.
Given the two lines AB and AC, and BC their dif
ference. Construct the square AGFB upon AB, the
square A CKH upon A C, and the square CDEB upon
BC. Produce ED to meet AG in L. The dimensions
of the rectangles LGFE and HLDK are AB and AC,
and the square CDEB is the difference between the
whole figure and the sum of these rectangles.
This proves geometrically the algebraic formula
(a _ 6)2 = a2  2 ab + b^.
23. The difference between the squares constructed upon'
two lines is equivalent to the rectangle of the sum and differ
ence of these lines. / k
Given the squares ^BD J? and CBFG, constructed
upon AB and BC. The difference between these
squares is the polygon ACGFDE, which is composed
of the rectangles ACHE and GFDH. Produce AE
and CH to / and K respectively, making EI and HK
each equal to BC, and draw IK. The difference be ^
tween the squares ABDE and CBFG is then equiva
lent to the rectangle ACKI, with dimensions AB + BC, and AB
This proves geometrically the algebraic formula
a2 62 = (a + 6) (a 6).
. 1>
\
H
G
BC.
214 BOOK IV. PLANE GEOMETRY
Proposition XIV. Problem
344. To construct a square equivalent to the sum of
tivo given squai^es.
Given the two squares, i? and i?'.
Required to construct a square equivalent to E \ E'.
Construction. Construct the rt. Z.4. § 228
On the sides of Z .4, take AB, or c, equal to a side of R', and
AC, or b, equal to a side of R, and draw BC, or a.
Construct the square S, having a side equal to BC.
Then ^ is the square required. q.e.f.
Proof. a=Z;'6'2. §337
{The square on the hypotenuse of a rt. A is equivalent to the sum of
the squares on the other two sides.)
.'. S = R{R', by Ax. 9. q.e.d.
345. Corollary 1. To construct a square equivalent to the
difference of two given squares.
We may easily reverse the above construction by first dravi^ing .c.
then erecting a ± at ^, and then with a radius a fixing the point C.
346. Corollary 2. To construct a square equivalent to the
Slim of three given squares.
If a side of the third square is d, we may erect a perpendicular from
C to the line J5C, take CD equal to d, and join D and B.
Discussion. It is evident that we can continue this process indefi
nitely, and thus construct a square equivalent to the sum of any number
of given squares.
PKOBLEMS OF CONSTKUCTION
215
Proposition XV. Problem
347. To construct ci j^ohjgon similar to two given
similar polygons and equivalent to their sicrn.
R"
Given the two similar polygons R and i?'.
Required to construct a polygon similar to R and R\ and
equivalent to R \ R\
Construction. Construct the rt AO. § 228
Let s and s' be corresponding sides of R and R\
On the sides of /.O, take OX equal to s', and OY equal to s.
Draw XY, and take s" equal to XY.
Upon s", corresponding to s, construct W similar to 11. § 312
Then it" is the polygon required. q.e.f.
Proof. OF^ \'0X'^ = XY^. § 337
Putting for OY, OX, and XF their equals s, s', and s", we have
But
and
By addition,
R"~s"^''
R"~s"^'
R^R'
s'\
R'
1.
:. R" = R + PJ, by Ax. 3.
Ax. 9
§334
Ax. 1
Q.E.D.
216
BOOK IV. PLANE GEOMETRY
Pkoposition XVI. Problem
348. To construct a triangle equivalent to a given
polygon.
Given the polygon ABCDEF.
Required to construct a triangle equivalent to ABCDEF.
Construction. Let B, C, and D be any three consecutive
vertices of the polygon. Draw the diagonal DB.
From V draw a line II to DB. § 233
Produce AB to meet this line at Q, and draw DQ.
Again, draw EQ, and from D draw a line II to EQ, meeting
AB produced at R, and draw ER.
In like manner continue to reduce the number of sides of
the polygon until we obtain the A EPR.
Then A £JP/? is the triangle required. q.e.f.
Proof. The polygon AQDEF has one side less than the
polygon ABCDEF.
Furthermore, in the two polygons, the ipSii'tABDEF is common,
and the A BQD = A BCD. § 326
(For the base DB is common, and their vertices C and Q are in
the line CQ II to the base.)
.. AQDEF = ABCDEF. Ax. 1
In like manner it may be proved that
AREF= AQDEF, and EPR = AREF. Q.e.d.
PROBLEMS OF CONSTRUCTION
217
Proposition XVII. Problem
349. To construct a square equivalent to a given
parallelogram.
p
N
M
A h B
Given the parallelogram ABCD.
Required to construct a square equivalent to the EJABCD.
Construction. Upon any convenient line take NO equal to a,
and OM equal to b, the altitude and base respectively of O
ABCD.
Upon NM as a diameter describe a semicircle.
At O erect OP _L to NM, meeting the circle at P. § 228
Construct the square S, having a side equal to OP.
Then S is the square required. q.e.f.
Proof.
That is,
But
and
NO :OP = OP: OM.
.'. 0P^=N0X0M.
OF''=ah.
S = 0P\
§297
§261
Ax. 9
EJABCD = ab. §322
.. S=CJABCD, by Ax. 9. q.e.d.
350. Corollary 1. To construct a square equivalent to a
given triangle.
Take for a side of the square the mean proportional between the base
and half the altitude of the triangle.
351. Corollary 2. To construct a square equivalent to a
given polygon.
First reduce the polygon to an equivalent triangle, and then construct
a square equivalent to the triangle.
218 BOOK IV. PLANE GEOMETRY
Proposition XVIII. Problem
352. To construct a parallelogram equivalent to a
given square, and having the sum of its base and
altitude equal to a given line.
A Q B
Given the square S, and the line AB.
Required to construct a O equivalent to S^ with the sum of
its base and altitude equal to AB.
Construction. Upon AB as a diameter describe a semicircle.
At A erect AC A to AB and equal to a side of the given
square S. § 228
Draw CD II to AB, cutting the circle at P. § 233
Draw PQ _L to AB. § 227
Then any CJ, as P, having AQ for its altitude and QB for
its base is equiva
lent to .S.
Q.E.F.
Proof.
AQ:PQ = PQ: QB.
§297
.'.Pq''=AQx QB.
§261
Furthermore
PQ is II toC^.
§95
.\PQ = CA.
§127
.'.PQ^ = CA\
Ax. 5
.•..4QX QB=CA\
Ax. 8
But
P = AQx QB,
§322
and
S=zCA\
§320
.'. P=S, by Ax. 8.
Q.E.D.
Thus is solved geometrically the algebraic problem, given x + y = a,
xy = 6, to find x and y.
PKOBLEMS OF CONSTRUCTION
219
Pkoposition XIX. Problem
353. To construct a parallelogram equivalent to a
given square, and having the difference of its base and
altitude equal to a given line,
c
.E^
\B
Given the square S, and the line AB.
Required to construct a O equivalent to S, ivith the differ
ence of its base and altitude equal to AB.
Construction. Upon .45 as a diameter describe a circle.
From A draw A C\ tangent to the circle, and equal to a side
of the given square S.
Through the center of the circle draw CD intersecting the
circle at E and D.
Then any O, as P, having CD for its base and CE for its
altitude, is equivalent to aS. q. e. f.
Proof. CD : CA = CA : CE. § 302
.'.CA^ = CDxCE, §261
and the difference between CD and CE is the diameter of the
circle, that is, AB.
But P=CDx CE, §322
and S=CA''. §320
.'. P= S,hy Ax. 8. Q. E. D.
Thus is solved geometrically the algebraic problem, given x — y = a,
xy = &, to find x and y.
220 BOOK IV. PLANE GEOMETRY
Proposition XX. Problem
354. To construct a j)olygon similar to a given poly
gon and equivalent to another given polygon.
Given the polygons P and Q.
Required to construct a polygon similar to P and equiva
lent to Q.
Construction. Construct squares equivalent to P and Q, § 351
and let m and n respectively denote their sides.
Let s be any side of P.
Find s\ the fourth proportional to m^ n, and s. § 307
Upon s\ corresponding to s,
construct a polygon P' similar to the polygon P. § 312
Then P' is the polygon required. q.e.f.
Proof. Since m : n = s : s', Const.
.•.m^:7i' = s^:s'\ §270
But P = 711^, and Q = n^. Const.
.'.P:Q = s'':s'\ Ax. 9
But P'.P^ = s' : s'l § 334
{The areas of two similar polygons are to each other as the squares on
any two corresponding sides.)
,'. P:Q = P:P'. Ax. 8
.'.P'=Q. §263
.*. P', being similar to P, is the polygon required, q.e.d.
PROBLEMS OF CONSTRUCTION
221
Proposition XXI. Problem
355. To construct a square which shall have a given
ratio to a given square.
s
D
#^. /
Given the square S, and the ratio — .
m
Required to construct a square which shall he to S as n is
to m.
Construction. Take AB equal to a side of S, and draw /IF,
making any convenient angle with AB.
On A Y take AE equal to m units and EF equal to n units.
Draw EB.
From F draw a line II to EB, meeting AB produced at C. § 233
On .4 C as a diameter describe a semicircle.
At B erect BD J_ to ^ C, meeting the semicircle at D.
Then BD is a side of the square required
Proof. Denote AB hj a, BC hj b, and BD by x.
§228
Q.E.F.
Then
But
By inversion,
a : x^x :
b.
. a:b = a^
:x\
a\h — m
: n.
a^ :x^ = m
: n.
x^ :a^ = n:
Tfl.
§297
§271
§273
Ax. 8
§266
Hence the square on BD will have the same ratio to S as
n has to m. q.e.d.
222 BOOK IV. PLANE GEOMETRY
Proposition XXII. Problem
356. To construct a jjolycjon similar to a given j)oly
gon and having a given ratio to it.
Given the polygon P and the ratio — •
772
Required to construct a polygon similar to P, which shall he
to P as n is to m.
Construction. Let s be any side of P.
Draw a line ,s', such that the square on *•' shall be to the
square on ^ as t^ is to m. § 355
Upon *•' as a side corresponding to s construct the polygon
P' similar to P. § 312
{Upon a given line corresponding to a given side of a given polygon^
to construct a polygon similar to the given polygon.)
Then P' is the polygon required. q. e. f.
Proof. P':P = s"':s\ §334
( The areas of two similar polygons are to each other as the squares
on any two corresponding sides.)
But s'^ : s'^ — n: m. Const.
Therefore P^ : P = n : m, hj Ax. 8. q.e.d.
This problem enables us to construct a square that is twice a given
square or half a given square, to construct an equilateral triangle that
shall be any number of times a given equilateral triangle, and in general
to enlarge or to reduce any figure in a given ratio. An architect's drawl
ing, for example, might need to be enlarged so as to be double the area
of the original, and the scale could be found by this method.
EXERCISES 223
EXERCISE 55
Problems of Computation
1. The sides of a triangle are 0.7 in., 0.6 in., and 0.7 in.
respectively. Is the largest angle acute, right, or obtuse ?
2. The sides of a triangle are 5.1 in., 6.8 in., and 8.5 in.
respectively. Is the largest angle acute, right, or obtuse ?
3. Find the area of an isosceles triangle whose perimeter
is 14 in. and base 4 in. (One decimal place.)
4. Find the area of an equilateral triangle whose perimeter
is 18 in. (One decimal place.)
5. Find the area of a right triangle, the hypotenuse being
1.7 in. and one of the other sides being 0.8 in.
6. Find the ratio of the altitudes of two triangles of equal
area, the base of one being 1.5 in. and that of the other 4.5 in.
7. The bases of a trapezoid are 34 in. and 30 in., and the
altitude is 2 in. Find the side of a square having the same area.
8. What is the area of the isosceles right triangle in which
the hypotenuse is V2 ?
9. AVhat is the area of the isosceles right triangle in which
the hypotenuse is 7 V2 ?
10. If the side of an equilateral triangle is 2 V3, what is the
altitude of the triangle ? the area of the triangle ?
11. If the side of an equilateral triangle is 1 ft., what is the
area of the triangle ?
12. If the area of an equilateral triangle is 43.3 sq. in., what
is the base of the triangle ? (Take V3 = 1.732.)
13. The sides of a triangle are 2.8 in., 3.5 in., and 2.1 in.
respectively. Draw the figure carefully and see what kind of
a triangle it is. Verify this conclusion by applying a geometric
test, and find the area of the triangle.
224 BOOK lY. PLANE GEOMETRY
EXERCISE 56
Theorems
1. The area of a rhombus is equal to half the product of
its diagonals.
2. Two triangles are equivalent if the base of the first
is equal to half the altitude of the second, and the altitude
of the first is equal to twice the base of the second.
3. The area of a circumscribed polygon is equal to half the
product of its perimeter by the radius of the inscribed circle.
4. Two parallelograms are equivalent if their altitudes are
reciprocally proportional to their bases.
5. If equilateral triangles are constructed on the sides of a
right triangle, the triangle on the hypotenuse is equivalent to
the sum of the triangles on the other two sides.
6. If similar polygons are constructed on the sides of a
right triangle, as corresponding sides, the polygon on the
hypotenuse is equivalent to the sum of the polygons on the
other two sides.
Ex. 6 is one of the general forms of the Pythagorean Theorem.
7. If lines are drawn from any point within a parallelogram
to the four vertices, the sum of either pair of triangles with
parallel bases is equivalent to the sum of the other pair.
8. Every line drawn through the intersection of the diag
onals of a parallelogram bisects the parallelogram.
9. The line that bisects the bases of a trapezoid divides the
trapezoid into two equivalent parts.
10. If either diagonal of a trapezoid bisects it, the trapezoid
is a parallelogram.
11. The triangle formed by two lines drawn from the mid
point of either of the nonparallel sides of a trapezoid to the
opposite vertices is equivalent to half the trapezoid.
EXERCISES 225
EXERCISE 57
Problems of Construction
1. Given a square, to construct a square of half its area.
2. To construct a right triangle equivalent to a given
oblique triangle.
3. To construct a triangle equivalent to the sum of two
given triangles.
4. To construct a triangle equivalent to a given triangle,
and having one side equal to a given line.
5. To construct a rectangle equivalent to a given parallelo
gram, and having its altitude equal to a given line.
6. To construct a right triangle equivalent to a given tri
angle, and having one of the sides of the right angle equal to
a given line,
7. To construct a right triangle equivalent to a given tri
angle, and having its hypotenuse equal to a given line.
8. To divide a given triangle into two equivalent parts by
a line through a given point P in the base.
9. To draw from a given point P in the base ylJ5 of a tri
angle ABC 2^ line to AC produced, so that it may be bisected
by BC.
10. To find a point within a given triangle such that the
lines from this point to the vertices shall divide the triangle
into three equivalent triangles.
11. To divide a given triangle into two equivalent parts by
a line parallel to one of the sides.
12. Through a given point to draw a line so that the seg
ments intercepted between the point and perpendiculars drawn
to the line from two other given points may have a given ratio.
13. To find a point such that the perpendiculars from it to
the sides of a given triangle shall be in the ratio p, q, r.
226 BOOK IV. PLANE GEOMETRY
EXERCISE 58
Review Questions
1. What is meant by the area of a surface ? Illustrate.
2. What is the difference between equivalent figures and
congruent figures ?
3. State two propositions relating to the ratio of one
rectangle to another.
4. Given the base and altitude of a rectangle, how is the area
found ? Given the area and base, how is the altitude found ?
5. How do you justify the expression, " the product of two
lines " ? " the quotient of an area by a line " ?
6. Can a triangle with a perimeter of 10 in. have the same
area as one with a perimeter of 1 in. ? Is the same answer
true for two squares ?
7. Can a parallelogram with a perimeter of 10 in. have the
same area as a rectangle with a perimeter of 1 in. ? Is the
same answer true for two rectangles ?
8. Explain how the area of an irregular field with straight
sides may be found by the use of the theorems of Book IV.
9. A triangle has two sides 5 and 6, including an angle of
70°, and another triangle has two sides 2 and 7^, including an
angle of 70°. What is the ratio of the areas of the triangles ?
 10. Two similar triangles have two corresponding sides 5 in.
and 15 in. respectively. The larger triangle has how many
times the area of the smaller ?
11. Given the hypotenuse of an isosceles right triangle, how
do you proceed to find the area ?
12. Given three sides of a triangle, what test can you apply
to determine whether or not it is a right triangle ?
13. Suppose you wish to construct a square equivalent to a
given polygon, how do you proceed ?
BOOK V
REGULAR POLYGONS AND CIRCLES
357. Regular Polygon. A polygon that is both equiangular
and equilateral is called a regular polygon.
Familiar examples of regular polygons are the equilateral triangle and
the square.
It is proved in Prop. I (§ 362) that a circle may be circumscribed about,
and a circle may be inscribed in, any regular polygon, and that these
circles are concentric (§ 188) .
358. Radius. The radius of the circle cir
cumscribed about a regular polygon is called
the radius of the polygon.
In this figure r is the radius of the polygon.
359. Apothem. The radius of the circle inscribed in a regular
polygon is called the apothem of the polygon.
In the figure a is the apothem of the polygon. The apothem is evi
dently perpendicular to the side of the regular polygon (§ 185) .
360. Center. The common center of the circles circumscribed
about and inscribed in a regular polygon is called the center of
the polygon.
In the figure O is the center of the polygon.
361. Angle at the Center. The angle between the radii drawn
to the extremities of any side of a regular polygon is called the
angle at the center of the polygon.
In the figure m is the angle at the center of the polygon. It is evidently
subtended by the chord which is the side of the inscribed polygon.
227
228
BOOK V. PLANE GEOMETRY
Proposition I. Theorem
362. A circle may he circumscribed about, and a circle
may he inscribed in, any regular polygon.
Given the regular polygon ABCDE.
To prove that 1. a circle may he circumscribed about ABCDE ;
2. a circle may be inscribed in ABCDE.
Proof. 1. Let O be the center of the circle which may be
passed through the three vertices A, B, and C. § 190
Draw OA, OB, OC, OD.
Then OB = OC, §162
and AB = CD. §357
Furthermore Z CBA = Z DCB, i 357
and Z CBO = Z OCB. § 74
.\Z0BA = ZDC0. Ax. 2
..A OAB is congruent to AOCD. § 68
.'.OA=OD. §67
Therefore the circle that passes through A, B, C, passes also
through D.
In like manner it may be proved that the circle that passes
through B, C, and D passes also through E; and so on.
Therefore the circle described with as a center and OA as a
radius will be circumscribed about the polygon, by § 205. q. e. d.
REGULAR POLYGOKS AND CIRCLES 229
Proof. 2. Let O be the center of the circumscribed circle.
D
Since the sides of the regular polygon are equal chords of
the circumscribed circle, they are equally distant from the
center. § 178
Therefore the circle described with as a center, with the
perpendicular from O to a side of the polygon as a radius, will
be inscribed in the polygon, by § 205. q.e.d.
363. Corollary 1. The radius drawn to any vertex of a
regular polygon bisects the angle at the vertex.
364. Corollary 2. The angles at the center of any regular
polygon are equal, and each is supplementary to an interior
angle of the polygon.
For the angles at the center are corresponding angles of congruent
triangles. If M is the midpoint of AB^ then since the A MOB and OBM
are complementary what can we say of their doubles, AOB and CBA ?
365. Corollary 3. An equilateral polygon inscribed in a
circle is a regular polygon.
Why are the angles also equal ?
366. Corollary 4. An equiangular polygon circumscribed
about a circle is a regular polygon.
By joining consecutive points of contact of the sides of the polygon
can you show that certain isosceles triangles are congruent, and thus
prove the polygon equilateral ?
230 BOOK V. PLANE GEOMETRY
Proposition II. Theorem
367. If a circle is divided into any number of equal
arcs, the chords joining the successive points of division
form a regular inscribed polygon; and the tangents
drawn at the points of division form a regidar circum
scribed polygon.
Given a circle divided into equal arcs by ^, 5, C, Z), and E^ AB^
BCj CD, DE, and EA being chords, and PQ, QR, RS, ST, and TP
being tangents at B, C, D, E, and A respectively.
To prove that 1. ABODE is a regular polygon ;
2. PQRST is a regular polygon.
Proof. 1. Since the arcs are equal by construction,
.. AB = BC = CD = DE = EA. § 170
.'. ABODE is a regular polygon. § 365
{An equilateral polygon inscribed in a circle is a regular polygon.)
Proof. 2. AP = Z.Q = ZR = ZS = ZT. §221
{An Z formed by two tangents is measured by half the difference of
the intercepted arcs.)
.*. PQRST is a regular polygon. § 366
{An equiangular polygon circumscribed about a circle is a
regular polygon.) Q. E. D.
KEGULAR POLYGONS AND CIRCLES
231
368. Corollary 1. Taiigents to a circle at the vertices of
a regular inscribed polygon form a regular circumscribed poly
gon of the same number of sides.
369. Corollary 2. Tangents to a circle at the midpoints
of the arcs subtended by the sides of a regular inscribed
polygon form a regular circumscribed
polygon^ whose sides are parallel to the
sides of the inscribed polygon and whose
vertices lie on the radii (^produced^ of ^^
the inscribed polygon.
For two corresponding sides, AB and A'B'^
are perpendicular to OM (§§ 176, 185), and are
parallel (§ 95) ; and the tangents MB' and iVJB',
intersecting at a point equidistant from OM and ON (§ 192), intersect upon
the bisector of the Z MON {% 152) ; that is, upon the radius OB (§ 363).
370. Corollary 3. Lines draum from each
vertex of a regular polygon to the 7nidpoints
of the adjacent arcs subtended by the sides of
the polygon form a regular inscribed polygon
of double the number of sides.
371. Corollary 4. Tangents at the mid
points of the arcs between adjacent points of
contact of the sides of a regular circumscribed
polygon form a regular circumscribed polygon ^^
of double the number of sides. j
372. Corollary 5. The perimeter of a regular inscribed
polygon is less than that of a regular inscribed polygon of
double the number of sides; and the perimeter of a regular
circumscribed polygori is greater than that of a regular cir
cumscribed polygofi of double the number of sides.
232 BOOK V. PLANE GEOMETRY
EXERCISE 59
1. Find the radius of the square whose side is 5 in.
2. Find the side of the square whose radius is 7 in.
3. Find the radius of the equilateral triangle whose side
is 2 in. ^
4. Find the side of the equilateral triangle whose radius
is 3 in.
5. Find the apothem of the equilateral triangle whose side
is VS in.
6. Find the side of the equilateral triangle whose apothem
is 2 V3 in.
7. How many degrees are there in the angle at the center
of an equiangular triangle ? of a regular hexagon ?
8. Given an equilateral triangle inscribed in a circle, to
circumscribe an equilateral triangle about the circle.
9. Given an equilateral triangle inscribed in a circle, to in
scribe a regular hexagon in the circle, and to circumscribe a
regular hexagon about the circle.
10. Given a square inscribed in a circle, to inscribe a regular
octagon in the circle, and to circumscribe a regular octagon
about the circle.
11. How many degrees are there in the angle at the center
of a regular octagon ? in each angle of a regular octagon ? in
the sum of these two angles ?
12. What is the area of the square inscribed in a circle of
radius 2 in.?
13. The apothem of an equilateral triangle is equal to half
the radius.
14. Prove that the apothem of an equilateral triangle is equal
to one fourth the diameter of the circumscribed circle. From
this show how an equilateral triangle may be inscribed in a circle.
REGULAE POLYGONS AND CIRCLES
Proposition III. Theorem
233
373. Tivo regular polygons of the same numher of
sides are similar.
§145
Given the regular polygons P and P', each having n sides.
To prove that P and P' are similar.
2(712)
Proof. Each angle of either polygon = rt. A.
2 (n — 2)
{Each Z of a regular polygon of n sides is equal to — rt. A.)
Hence the polygons P and P' are mutually equiangular.
Furthermore, '.' AB = BC = CD = DE = EA,
and A'B' = B'C = CD' = D'E' = E'A', § 357
AB _ BC _ CD _ DE _ EA
' ' A'B'~ B'C'~ CD' ~ D'E' ~ E'A''
Hence the polygons have their corresponding sides propor
tional and their corresponding angles equal.
Therefore the two polygons are similar, by § 282. q.e.d.
374. Corollary. The areas of two regular polygons of the
same numher of sides are to each other as the squares on any
two corresponding sides.
Por the areas of two similar polygons are to each other as tlie squares
on any two corresponding sides (§ 334), and two regular polygons of the
same number of sides are similar.
Ax. 4
234
BOOK V. PLANE GEOMETEY
Proposition IV. Theorem
375. The perimeters of two regular p)olygons of the
same number of sides are to each other as their radii,
and also as their apothems.
A MB A' M' B'
Given the regular polygons with perimeters p and p\ radii r
and r', apothems a and a', and centers O and 0' respectively.
To prove that p : p' = r : r' = a: a'.
Proof. Since the polygons are similar,
.\p:2)' = AB:A'B'.
Furthermore in the isosceles AOAB and O'A'B',
and OA : OB = O'A' : O'B'.
{For each of these ratios equals 1.)
..the AOAB and O'A'B' are similar.
.'.AB:A'B' = r:r'.
Also A A MO and A'M'O' are similar.
.'. r :r' = a : a'.
.. j9 i]^' = r : r' = a : a', by Ax. 8.
376. Corollary. The areas of two regular polygons of the
same number of sides are to each other as the squares on the
radii of the circumscribed circles, and also as the squares on
the radii of the inscribed circles.
§373
§291
§364
Ax. 8
§288
§282
§286
§282
Q. E. D.
KEGULAR POLYGONS AND CIRCLES 235
EXERCISE 60
1. Eind the ratio of the perimeters and the ratio of the
areas of two regular hexagons, their sides being 2 in. and
4 in. respectively.
2. Find the ratio of the perimeters and the ratio of the
areas of two regular octagons, their sides having the ratio 2 : 6.
3. Eind the ratio of the perimeters of two squares whose
areas are 121 sq. in. and 30i sq. in. respectively.
4. Eind the ratio of the perimeters and the ratio of the
areas of two equilateral triangles whose altitudes are 3 in.
and 12 in. respectively.
5. The area of one equiangular triangle is nine times that
of another. Required the ratio of their altitudes.
6. The area of the cross section of a steel beam 1 in. thick
is 12 sq. in. What is the area of the cross section of a beam of
the same proportions and 1^ in. thick ?
7. Squares are inscribed in two circles of radii 2 in. and
6 in. respectively. Eind the ratio of the areas of the squares,
and also the ratio of the perimeters.
8. Squares are inscribed in two circles of radii 2 in. and
8 in. respectively, and on the sides of these squares equi
lateral triangles are constructed. What is the ratio of the
areas of these triangles ?
9. A round log a foot in diameter is sawed so as to have
the cross section the largest square possible. What is the area
of this square ? What would be the area of the cross section
of the square beam cut from a log of half this diameter ?
10. Every equiangular polygon inscribed in a circle is regular
if it has an odd number of sides.
11. Every equilateral polygon circumscribed about a circle
is regular if it has an odd number of sides.
236 BOOK V. PLANE GEOMETRY
Proposition V. Theorem
377. If the number of sides of a regular inscribed
polygon is indefinitely increased, the ai^otheni of the
polygon approaches the radius of the circle as its limit.
Given a regular polygon of n sides inscribed in the circle of
radius OA^ s being one side and a the apothem.
To prove that a approaches r as a limits if n is increased
indefinitely.
Proof. We know that a<r. § 86
Then since r~a<AM, § 112
and AM<s, §174
.\r — a<,s. Ax. 10
If n is taken sufficiently great, s, and consequently r— a, can
be made less than any assigned positive value, however small.
Since r—a can become and remain less than any assigned
positive value by increasing n, it follows that
r is the limit of a, by § 204. q.'e.d.
378. Corollary. // the number of sides of a regular in
scribed polygon is indefinitely increased^ the square on the
apothem approaches the square on the radius of the circle
as its limit.
For r2  a2 ^ AM'^. § 338
Therefore by taking n sufficiently great, s, and consequently AM, and
consequently r^ — a^, approaches zero as its limit.
REGULAR POLYGONS AND CIRCLES 237
Proposition VI. Theorem
379. An arc of a circle is less than a line of any kind
that envelops it on the convex side and has the same
extremities.
Given BCA an arc of a circle, AB being its chord.
To prove that the arc BCA is less than a line of any hind
that envelops this arc and terminates at A and B.
Proof. Of all the lines that can be drawn, each to include
the area ABC between itself and the chord AB, there must be
at least one shortest line ; for all the lines are not equal.
Let BDA be any kind of line enveloping BCA as stated.
The enveloping line BDA cannot be the shortest ; for draw
ing ECF tangent to the arc BCA at any point C, the line
BFCEA<BFDEA, since FCE< FDE. Post. 3
In like manner it can be shown that no other enveloping
line can be the shortest.
.*. BCA is shorter than any enveloping line. q.e.d.
380. Corollary. A circle is less than the perimeter of any
polygon circumscribed about it.
381. Circle as a Limit. From Prop. VI we may now assume :
1. The circle is the limit which the perimeters of regular in
scribed polygons and of similar circumscribed polygons approach,
if the number of sides of the polygons is indefinitely increased.
2. The area of the circle is the limit which the areas of the
inscribed and circumscribed polygons approach.
238
BOOK V. PLANE GEOMETEY
Proposition YII. Theorem
382. Tico circumferences hcwe the same ratio as their
radii.
Given the circles with circumferences c and c', and radii r and
r' respectively.
To prove that c: c' = r : r'.
Proof. Inscribe in the circles two similar regular polygons,
and denote their perimeters by ^ and^>'.
Then p:2)' = r:7''. §375
Conceive the number of sides of these regular polygons to
be indefinitely increased, the polygons continuing similar.
Thenp and 2^' will have c and c' as limits. § 381
But the ratio p :p' will always be equal to the ratio ?• : r'
.'. jyr' =2)'t'.
.'. cr' = c'r.
,'.c:c' = r:r', by §264.
383. Corollary 1. The ratio of any circle to its diameter
is constant.
Why does c : c'= 2 r : 2 r' ? Then why does c : 2 r = c' : 2 r' ?
384. The Symbol tt. The constant ratio of a circle to its
diameter is represented by the Greek letter rrr (pi).
385. Corollary 2. In any circle c=2irr.
§375
§261
§207
Q.E.D.
For, by definition, 17 =
2r
REGULAR POLYGONS AND CIRCLES 239
Proposition VIII. Theorem
386. The area of a regular polygon is equal to half
the product of its apothem hy its perimeter.
A M B
Given the regular polygon ABCDEF, with apothem a, perimeter />,
and area s.
To prove that * — 2 ^P'
Proof. Draw the radii OA, OB, OC, etc., to the successive
vertices of the polygon.
The polygon is then divided into as many triangles as it
has sides.
The apothem is the common altitude of these A, and the
area of each A is equal to \a multiplied by the base. § 325
Hence the area of all the triangles is equal to \ a multiplied
by the sum of all the bases. Ax. 1
But the sum of the areas of all the triangles is equal to the
area of the polygon. Ax. 11
And the sum of all the bases of the triangles is equal to the
perimeter of the polygon. Ax. 11
.. s = ^ ap. Q.E. D.
387. Similar Parts. In different circles similar arcs, similar
sectors, and similar segments are such arcs, sectors, and seg
ments as correspond to equal angles at the center.
For example, two arcs of 30° in different circles are similar arcs, and
the corresponding sectors are similar sectors.
240 BOOK V. PLANE GEOMETRY
Proposition IX. Theorem
388. The area of a circle is equal to half the product
of its radius hy its circumference.
Given a circle with radius r, circumference c, and area s.
To prove that
s = 1 re.
Proof. Circumscribe any regular polygon of n sides, and
denote the perimete* of this polygon by p and its area by s\
Then since r is its apothem, s' = ^ 77^. § 386
Conceive n to be indefinitely increased.
Then since p approaches c as its limit, § 381
and r is constant,
'■ \rp approaches \rc as its limit.
Also s' approaches s as its limit. § 381
But s' = \rp always. § 386
.\s = lrc, by § 207. q.e.d.
389. Corollary 1. The area of a circle is equal to ir times
the square on its radius.
For the area of the O = \rc = \r x ^irr — tit^.
390. Corollary 2. The areas of two circles are to each other
as the squares on their radii.
391. Corollary 3. The area of a sector is equal to half the
product of its radius hy its are.
area of sector arc of sector
For
area of circle circle
EEGULAR POLYGONS AND CIRCLES 241
EXERCISE 61
1. Two circles are constructed with radii 1^^ in. and 4^ in.
respectively. The circumference of the second is how many
times that of the first?
2. The circumference of one circle is three times that of
another. The square on the radius of the first is how many
times the square on the radius of the second?
3. The circumference of one circle is 2^ times that of
another. The equilateral triangle constructed on the diameter
of the first has how many times the area of the equilateral
triangle constructed on the diameter of the second ?
4. A circle with a diameter of 5 in. has a circumference of
15.708 in. What is the circumference of a pipe that has a
diameter of 2 in. ?
5. A wheel with a circumference of 4 ft. has a diameter
of 1.27 ft., expressed to two decimal places. What is the cir
cumference of a wheel with a diameter of 1.58 ft. ?
6. A regular hexagon is 2 in. on a side. Find its apothem
and its area to two decimal .places.
7. An equilateral triangl^is 2 in. on a side. Find its apothem
and its area to two decimal places.
8. The radius of one circle is 2^ times that of another.
The area of the smaller is 15.2 sq. in. What is the area of
the larger ?
9. The radius of one circle is 3^ times that of another.
The area of the smaller is 17.75 sq. in. What is the area of
the larger ?
10. The circumferences of two cylindrical steel shafts are
respectively 3 in. and 1^ in. The area of a cross section of the
first is how many times that of a cross section of the second ?
11. The arc of a sector of a circle 2i in. in diameter is 1 in.
What is the area of the sector ?
242 BOOK V. PLANE GEOMETRY
Proposition X. Problem
392. To inscribe a square in a given circle.
Given a circle with center O.
Required to inscribe a square in the given circle.
Construction. Draw two diameters A C and BD perpendicular
to each other. § 228
Draw AB, BC, CD, and DA.
Then A BCD is the square required.
Proof. The A CBA, DCB, ADC, BAD are rt. A.
{An Z inscribed in a semicircle is a rt. Z.)
The A at the centef being rt. A,
the arcs AB, BC, CD, and DA are equal,
and the sides AB, BC, CD, and DA are equal.
Hence the quadrilateral A BCD is a square, by § 65. q.e.d.
393. Corollary. To inscribe regular polygons of ^,1Q,Z2,
64, etc., sides in a given circle.
By bisecting the arcs AB., BC, etc., a regular polygon of how many
sides may be inscribed in the circle ? By continuing the process regular
polygons of how many sides may be inscribed ?
In general we may say that this corollary allows us to inscribe a reg
ular polygon of 2« sides, where n is any positive integer. As a special
case it is interesting to note that n may equal 1.
Q.E.F.
§215
Const.
§212
§170
PEOBLEMS OF CONSTRUCTION 243
Proposition XI. Problem
394. To inscribe a regular hexagon in a given circle.
A>~ — ^B
Given a circle with center 0.
Required to inscribe a regular hexagon in the given circle.
Construction. From the center draw any radius, as OC.
With C as a center, and a radius equal to OC, describe an
arc intersecting the circle at D.
Draw OD and CD.
Then CD is a side of the regular hexagon required, and
therefore the hexagon may be inscribed by applying CD six
times as a chord. q.e.f.
Proof. The A OCD is equiangular. § 75
{An equilateral triangle is equiangular.)
Hence the Z COD is ^ of 2 rt. Zs, or ^ of 4 rt. A. § 107
.'. the arc CD is ^ of the circle.
.'. the chord CD is a side of a regular inscribed hexagon, q. e. d.
395. Corollary 1. To inscribe an equilateral triangle in
a given circle.
By joining the alternate vertices of a regular inscribed hexagon, an
equilateral triangle may be inscribed.
396. Corollary 2. To inscribe regular polygons of 12, 24,
48, etc., sides in a given circle.
244 BOOK V. PLANE GEOMETRY
Proposition XII. Problem
397. To inscribe a regular decagon in a given circle.
Given a circle with center O.
Required to inscribe a regular decagon in the given circle.
Construction. Draw any radius OA,
and divide it in extreme and mean ratio,
so that OA:OP=OP:AP. § 311
From A as a center, with a radius equal to OP,
describe an arc intersecting the circle at B.
Draw AB.
Then A Bis a. side of the regular decagon required, and there
fore the regular decagon may be inscribed by applying AB ten
Q. E. F.
Draw PB and OB.
OA : OP = OP : APj
AB = OP.
'.OA:AB = AB:AP.
ZBAO = ZBAP.
Hence the AOAB and BAP are similar.
But the AOAB is isosceles.
.*. A BAP, which is similar to AOAB, is isosceles, '§ 282
and AB = BP = OP. §62
times as a chord
Proof.
Then
and
Moreover,
Const.
Const.
Ax. 9
Iden.
§288
§162
PROBLEMS OF CONSTRUCTION .245
The A PBO being isosceles, the Z O = Z OBP. § 74
But the Z^P5 = Z0hZ0^P = 2Z0. §111
Hence ABAP = 2/.0,
and ^ ZOi5^ = 2ZO. Ax. 9
..the sum of the A of the A OAB = 5 Z O = 2 rt. Z, § 107
and Z O = 1 of 2 rt. A, or ^^ of 4 rt. A. Ax. 4
Therefore the arc ^jB is j^^ of the circle. § 212
.'. the chord ^5 is a side of a regular inscribed decagon. q.e.d.
398. Corollary 1. To inscribe a regular pentagon in a
given circle.
399. Corollary 2. To inscribe regular polygons of 20, 40,
80, etc.^ sides in a given circle.
By bisecting the arcs subtended by the sides of a regular inscribed
decagon a regular polygon of how many sides may be inscribed ? By con
tinuing the process regular polygons of how many sides may be inscribed ?
EXERCISE 62
If r denotes the radius of a regular inscribed polygon, a
the apothem, s one side, A an angle, and C the angle at the
center, show that:
1. In a regular inscribed triangle s = r Vs, a = ^r, A = 60°,
C=120°.
2. In a regular inscribed quadrilateral s = r V2, a = ^r V2,
A = 90°, C = 90°.
3. In a regular inscribed hexagon s = r,a= ^r Vs, A = 120°,
C = 60°.
4. In a regular inscribed decagon
s = ^^ 51) ^ ^ ^ 1^ VlO + 2 V5, ^ =144°, C= 36°.
246 BOOK V. PLANE GEOMETRY
Proposition XIII. Problem
400. To inscribe in a given circle a recjulcir jpentadec
agon, or polygon of fifteen sides.
Given a circle.
Required to inscribe a regular pentadecagon in the given
circle.
Construction. Draw a chord PH equal to the radius of the
circle, a chord PA equal to a side of the regular inscribed
decagon, and draw AB.
Then ^5 is a side of the regular pentadecagon required, and
therefore the regular pentadecagon may be inscribed by apply
ing AB fifteen times as a chord. q.e.f.
Proof. The arc PB is i of the circle, § 394
and the arc PA is ^^ of the circle. § 397
Hence the arc AB \?> ^ — J^j, or ^^, of the circle. Ax. 2
.*. the chord .45 is a side of the regular inscribed pentadec
agon required. q.e.d.
401. Corollary. To inscribe regular polygons (^f 30, 60,
120, etc.^ sides in a given circle.
By bisecting the arcs AB^ BC, etc., a regular polygon of how many
sides may be inscribed ? By continuing the process regular polygons of
how many sides may be inscribed ? In general we may say that a regu
lar polygon of 15 • 2« sides may be inscribed in this manner.
PKOBLEMS OF CONSTRUCTION 247
EXERCISE 63
1. A fivecent piece is placed on the table. How many five
cent pieces can be placed around it, each tangent to it and
tangent to two of the others ? Prove it.
2. What is the perimeter of an equilateral triangle inscribed
in a circle with radius 1 in. ?
3. What is the perimeter of an equilateral triangle circum
scribed about a circle with radius 1 in. ?
4. What is the perimeter of a regular hexagon circumscribed
about a circle with radius 1 in, ?
Required to circumscribe about a given circle the following
regular polygons :
5. Triangle. 7. Hexagon. 9. Pentagon.
6. Quadrilateral. 8. Octagon. 10. Decagon.
11. Required to describe a circle whose circumference equals
the sum of the circumferences of two circles of given radii.
12. Required to describe a circle whose area equals the sum
of the areas of two circles of given radii.
13. Required to describe a circle having three times the area
of a given circle.
Required to construct an angle of :
14. 18°. 15. 36°. 16. 9°. 17. 12°. 18. 24°.
Required to construct tvith a side of given length :
19. An equilateral triangle. 23. A regular pentagon.
20. A square. 24. A regular decagon.
21. A regular hexagon. 25. A regular dodecagon.
22. A regular octagon. 26. A regular pentadecagon.
27. From a circular log 16 in. in diameter a builder wishes
to cut a column with its cross section as large a regular octagon
as possible. Find the length of each side.
248 BOOK V. PLANE GEOMETRY
Proposition XIV. Problem
402. Given the side and the radius of a regular in
scribed polygon, to find the side of the regular inscribed
polygon of double the nuviber of sides.
Given AB, the side of a regular inscribed polygon of radius OA.
Required to find AP, a side of the regular inscribed poly
yon of double the number of sides.
Solution. Denote the radius by r, and the side AB by s.
Draw the diameter PQ 1. to AB, and draw AO and AQ.
Then AM=\s.
In the rt. A. 4 03/, oFf = 7^\s^.
Therefore OM = ^r  i s\
Since PM^OM=r,
therefore PM =r— OM
Furthermore
= r — V/*^ — 1 s^.
AP =PQX PM.
But PQ = 2r, and PM =r — Vr^
s\
.'. AP = 2 r(r  V?"  i s%
.'.AP
= Vr(2r V4r^6').
§174
§338
Ax. 5
Ax. 11
Ax. 2
Ax. 9
§298
Ax. 9
Ax. 5
Q. E. F.
403. Corollary. 7/ r =1, ^P = \/2  V4  s'
PROBLEMS OF COMPUTATION
Proposition XV. Problem
249
404. To find the numerical value of the ratio of the
circmnference of a circle to its diameter.
Given a circle of circumference c and diameter d.
c
Required to find the numerical value of  or ir.
ct
Solution. By § 385, 2 7rr = c. . • , tt = ^ c when r = 1.
Let Sg (read " s sub six ") be the length of a side of a regular
polygon of 6 sides, s^^ of 12 sides, and so on.
If r = 1, by § 394, ^<?g = 1, and by § 403 we have
Form of Computation
= \/2V4l^
s^ = V2  V4^0.51763809)2
s^3 = V2  V4^ (0.26105238^
s^ = V2  Vr (0.13080626)^
^m = V2  V4  (0.06543817)^
s^^ = V2 V4  (0.03272346)2
s_g3 = V2  V4^(0.01636228)^
.'.c = 6.28317 nearly ; that is^
TT is an incommensurable number. We generally take
TT = 3.1416, or ^, and  = 0.31831.
Length of Side
Length
of Perimeter
0.51763809
6.21165708
0.26105238
6.26525722
0.13080626
6.27870041
0.06543817
6.28206396
0.03272346
6.28290510
0.01636228
6.28311544
0.00818121
6.28316941
= 3.14159 nearly. Q.e.f.
250 BOOK V. PLAKE GEOMETRY
EXERCISE 64
Problems of Computation
Using the value S.1416 for tt, find the cireumferences of
circles with radii as follows :
1. 3 in. 3. 2.7 in. 5. 7 in. 7. 2 ft. 8 in.
2. 5 in. 4. 3.4 in. 6. 6 in. 8. 3 ft. 7 in.
Find the circumferences of circles with diameters as follows :
9. 9 in. 11. 5.9 in. 13. 2^ ft. 15. 29 centimeters.
10. 12 in. 12. 7.3 in. 14. 3^ in. 16. 47 millimeters.
Find the radii of circles with circumferences as follows :
17. Tit. 19. 15.708 in. 21. 18.8496 in. 23. 345.576 ft.
18. 3^77. 20. 21.9912 in. 22. 125.664 in. 24. 3487.176 in.
Find the diameters of circles with circumferences as follows :
25. 15 TT. 27. 2 7rr. 29. 188.496 in. 31. 3361.512 in.
26. 7r\ 28. 7 7ral 30. 219.912 in. 32. 3173.016 in.
Find the areas of circles ivith radii as folloivs :
33. 5 X. 35. 27 ft. 37. 3^ in. 39. 2 ft. 6 in.
34. 2 TT. 36. 4.8 ft. 38. 4f in. 40. 7 ft. 9 in.
Find the areas of circles with diameters as follows :
41. 16aZ». 43.2.5 ft. 45. 3 yd. 47. 3 ft. 2 in.
42. 24 tt". 44. 7.3 in. 46. 4f yd. 48. 4 ft. 1 in.
Find the areas of circles tvith circumferences as folloivs :
49. 2 7r. 51. ira. 53. 18.8496 in. 55. 333.0096 in.
50. 4 7r. 52. 14 Tra^. 54. 329.868 in. 56. 364.4256 in.
Find the radii of circles with areas as follows :
57. na^*. 59. tt. 61. 12.5664. 63. 78.54.
58. 4 7rmV. 60. 2 7r. 62. 28.2744. 64. 113.0976.
EXEKCISES 251
EXERCISE 65
Problems of Construction
1. To inscribe in a given circle a regular polygon similar to
a given regular polygon.
2. To divide by a concentric circle the area of a given circle
into two equivalent parts.
3. To divide by concentric circles the area of a given circle
into n equivalent parts.
4. To describe a circle whose circumference is equal to the
difference of two circumferences of given radii.
5. To describe a circle the ratio of whose area to that of
a given circle shall be equal to the given ratio m, : n.
6. To construct a regular pentagon, given one of the
diagonals.
7. To draw a tangent to a given circle such that the seg
ment intercepted between the point of contact and a given
line shall have a given length.
8. In a given equilateral triangle to inscribe three equal
circles tangent each to the other two, each circle being tangent
to two sides of the triangle.
9. In a given square to inscribe four equal circles, so that
each circle shall be tangent to two of the others and also
tangent to two sides of the square.
10. In a given square to inscribe four equal circles, so that
each circle shall be tangent to two of the others and also
tangent to one side and only one side of the square.
11. To draw a common secant to two given circles exterior
to each other, such that the intercepted chords shall have the
given lengths a and b.
12. To draw through a point of intersection of two given
intersecting circles a common secant of a given length.
252 BOOK V. PLANE GEOMETRY
EXERCISE 66
Problems of Loci
1. Find the locus of the center of the circle inscribed in a
triangle that has a given base and a given angle at the vertex.
2. Find the locus of the intersection of the perpendiculars
from the three vertices to the opposite sides of a triangle that
has a given base and a given angle at the vertex.
3. Find the locus of the extremity of a tangent to a given
circle, if the length of the tangent is equal to a given line.
4. Find the locus of a point from which tangents drawn
to a given circle form a given angle.
5. Find the locus of the midpoint of a line drawn from
a given point to a given line.
6. Find the locus of the vertex of a triangle that has a
given base and a given altitude.
7. Find the locus of a point the sum of whose distances
from two given parallel lines is constant.
8. Find the locus of a point the difference of whose dis
tances from two given parallel lines is constant.
9. Find the locus of a point the sum of whose distances
from two given intersecting lines is constant.
10. Find the locus of a point the difference of whose dis
tances from two given intersecting lines is constant.
11. Find the locus of a point whose distances from two
given points are in the ratio m : n.
12. Find the locus of a point whose distances from two
given parallel lines are in the ratio m : n.
13. Find the locus of a point whose distances from two
given intersecting lines are in the ratio m:n.
14. Find the locus of a point the sum of the squares of
whose distances from two given points is constant.
EXERCISES 253
EXERCISE 67
Examination Questions
1. Each side of a triangle is 2 ti centimeters, and about each
vertex as a center, with a radius of n centimeters, a circle is
described. Find the area bounded by the three arcs that lie
outside the trid:ngle, and the area bounded by the three arcs
that lie within the triangle.
2. Upon a line AB 2i segment of a circle containing 240° is
constructed, and in the segment any chord CD subtending an
arc of 60° is drawn. Find the locus of the intersection of ^C
and J5Z), and also of the intersection oi AD and BC.
3. Three successive vertices of a regular octagon are A, B,
and C. If the length AB is a, compute the length A C.
4. The areas of similar segments of circles are proportional
to the squares on their radii.
5. An arc of a certain circle is 100 ft. long and subtends
an angle of 25° at the center. Compute the radius of the circle
correct to one decimal place.
6. Given a circle whose radius is 16, find the perimeter and
the area of the regular inscribed octagon.
7. If two circles intersect at the points A and B, and through
A a variable secant is drawn, cutting the circles in C and D, the
angle CBD is constant for all positions of the secant.
8. If A and B are two fixed points on a given circle, and P
and Q are the extremities of a variable diameter of the same
circle, find the locus of the point of intersection of the lines
AP and BQ.
9. The radius of a circle is 10 ft. Two parallel chords are
drawn, each equal to the radius. Find that part of the area of
the circle lying between the parallel chords.
The propositions in Exercise 67 are taken from recent college entrance
examination papers.
254 BOOK V. PLANE GEOMETRY
EXERCISE 68
Formulas
If r denotes the radius of a circle^ and s one side of a reg
ular inscribed polygon, prove the following, and find the value
of s to two decimal places when r —1 :
1. In an equilateral triangle s = r VS.
2. In a square s = r V2.
3. In a regular pentagon 5 = ^ ?• V 10 — 2 V 5.
4. In a regular hexagon s = r.
5. In a regular octagon s = r v2 — V2.
6. In a regular decagon s = ^ r (Vs — 1).
7. In a regular dodecagon s = r\2 — V 3.
8. A regular pentagon is inscribed in a circle whose radius
is r. If the side is s, find the apothem.
9. A regular polygon is inscribed in a circle whose radius
is r. If the side is s, show that the apothem is \ V4 7^ — s\
10. If the radius of a circle is r, and the side of an inscribed
regular polygon is s, show that the side of the similar cir
cumscribed regular polygon is
11. Three equal circles are described, each tangent to the
other two. If the common radius is r, find the area contained
between the circles.
12. Given p, P, the perimeters of regular polygons of n sides
inscribed in and circumscribed about a given circle. Eind p',
P', the perimeters of regular polygons of 2n sides inscribed in
and circumscribed about the given circle.
13. A circular plot of land d ft. in diameter is surrounded
by a walk w ft. wide. Find the area of the circular plot and
the area of the walk.
EXERCISES 255
EXERCISE 69
Applied Problems
1. The diameter of a bicycle wheel is 28 in. How many
revolutions does the wheel make in going 10 mi. ?
2. Find the diameter of a carriage wheel that makes 264
revolutions in going half a mile.
3. A circular pond 100 yd. in diameter is surrounded by a
walk 10 ft. wide. Find the area of the walk.
4. The span (chord) of a bridge in the form of a circular
arc is 120 ft., and the highest point of the arch is 15 ft. above
the piers. Find the radius of the arc.
5. Two branch water pipes lead into a main pipe. It is
necessary that the crosssection area of the main pipe shall
equal the sum of the cross sections of the two branch pipes.
The diameters of the branch pipes are respectively 3 in. and
4 in. Required the diameter of the main pipe.
6. A kite is made as here shown, the semicircle / , \
having a radius of 9 in., and the triangle a height
of 25 in. Find the area of the kite.
7. In making a drawing for an arch it is required
to mark off on a circle drawn with a radius of 5 in.
an arc that shall be 8 in. long. This is best done by
finding the angle at the center. How many degrees are there in
this angle ?
8. In an iron washer here shown, the diameter
of the hole is 1 in. and the width of the washer
is I in. Find the area of one face of the washer.
9. Find the area of a fan that opens out into a sector of
120°, the radius being 9 in.
10. The area of a fan that opens out into a sector of 111° is
96.866 sq. in. What is the radius ? (Take tt = 3.1416.)
256 BOOK V. PLANE GEOMETRY
EXERCISE 70
Review Questions
1. What is meant by a regular polygon ? by its radius ?
by its center? by its apothem ?
2. What other names are there for a regular triangle and
a regular quadrilateral ?
3. If one angle of a regular polygon is known, how can
the number of sides be determined?
4. The sides of two regular polygons of n sides are respec
tively s and s\ What is the ratio of their radii ? of their
apothems ? of their perimeters ? of their areas ?
5. The diameters of two circles are d and d' respectively.
What is the ratio of their radii ? of their circumferences ? of
their areas ?
6. If the number of sides of a regular inscribed polygon
is indefinitely increased, what is the limit of the apothem?
of each side ? of the perimeter ? of the area ? of the angle
at the center ? of each angle of the polygon ?
7. How do you find the area of a regular polygon ? of an
irregular polygon ? of a square ? of a triangle ? of a parallelo
gram ? of a circle ? of a trapezoid ? of a sector ?
8. What regular polygons have you learned to inscribe in
a circle ? Name three regular polygons that you have not
learned to inscribe.
9. Given the circumference of a circle, how can the area of
the circle be found ?
10. Given the area of a circle, how can the circumference of
the circle be found ?
11. What is the radius of the circle of which the number of
linear units of circumference is equal to the number of square
units of area ?
EXEECISES 257
EXERCISE 71
General Review of Plane Geometry
Write a classification of the different kinds of:
1. Lines. 3. Triangles. 5. Polygons.
2. Angles. 4. Quadrilaterals. 6. Parallelograms.
State the conditions under which :
7. Two triangles are congruent.
8. Two parallelograms are congruent.
9. Two triangles are similar.
10. Two straight lines are parallel.
11. Two parallelograms are equivalent.
12. Two polygons are similar.
Complete the following statements in the most general manner:
13. In any triangle the square on the side opposite • • •.
14. If two parallel lines are cut by a transversal, • • •.
15. If four quantities are in proportion, they are in pro
portion by • • • .
16. If two secants of a circle intersect, the angle formed is
measured by •••.
17. The perimeters of two similar polygons are to each
other as • • • .
18. The areas of two similar polygons are to each other as • • •.
19. The area of a circle is equal to • • •.
20. In the same circle or in equal circles equal chords • • •.
21. In the same circle or in equal circles the central angles
subtended by two arcs are • • • .
22. If two secants of a circle intersect within, on, or outside
the circle, the product of • • • .
258 BOOK V. PLANE GEOMETRY
23. If four lines meet in a point so that the opposite angles
are equal, these lines form two intersecting straight lines.
24. If squares are constructed outwardly on the six sides
of a regular hexagon, the exterior vertices of these squares are
the vertices of a regular dodecagon.
25. In a right triangle the line joining the vertex of the
right angle to the midpoint of the hypotenuse is equal to half
the hypotenuse.
26. No two lines drawn from the vertices of the base angles
of a triangle to the opposite sides can bisect each other.
27. The rhombus is the only parallelogram that can be cir
cumscribed about a circle.
28. The square is the only rectangle that can be circum
scribed about a circle.
29. No oblique parallelogram can be inscribed in a circle.
30. If two triangles have equal bases and equal vertical
angles, the two circumscribing circles have equal diameters.
31. If the inscribed and circumscribed circles of a triangle
are concentric, the triangle is equilateral.
32. If the three points of contact of a circle inscribed in a tri
angle are joined, the angles of the resulting triangle are all acute.
33. The diagonals of a regular pentagon intersect at the
vertices of another regular pentagon.
34. If two perpendicular radii of a circle are produced to
intersect a tangent to the circle, the other tangents from the
two points of intersection are parallel.
35. The line that joins the feet of the perpendiculars drawn
from the extremities of the base of an isosceles triangle to the
equal sides is parallel to the base.
36. The sum of the perpendiculars drawn to the sides of a
regular polygon from any point within the polygon is equal
to the apothem multiplied by the number of sides.
EXERCISES 259
37. If two consecutive angles of a quadrilateral are right
angles, the bisectors of the other two angles are perpendicular.
38. If two opposite angles of a quadrilateral are right angles,
the bisectors of the other two angles are parallel.
39. The two lines that join the midpoints of opposite sides
of a quadrilateral bisect each other.
40. The sum of the angles at the vertices of a fivepointed
star is equal to two right angles.
41. The segments of any line intercepted between two con
centric circles are equal.
42. The diagonals of a trapezoid divide each other into
segments which are proportional.
43. Given the midpoints of the sides of a triangle, to con
struct the triangle.
44. To divide a given triangle into two equivalent parts by
a line through one of the vertices.
45. To draw a tangent to a given circle that shall also be
perpendicular to a given line.
46. To divide a given line into two segments such that the
square on one shall be double the square on the other.
47. If any two consecutive sides of an inscribed hexagon
are respectively parallel to their opposite sides, the remaining
two sides are parallel.
48. If through any given point in the common chord of two
intersecting circles two other chords are drawn, one in each
circle, their four extremities will all lie on a third circle.
49. If two chords intersect at right angles within a circle,
the sum of the squares on their segments equals the square
on the diameter. Investigate the case in which the chords
intersect outside the circle ; also the case in which they inter
sect on the circle.
260 BOOK V. PLANE GEOMETRY
50. The lines bisecting any angle of an inscribed quadrilateral
and the opposite exterior angle intersect on the circle.
51. The sum of the perpendiculars from any point in an
equilateral triangle to the three sides is constant.
52. The perpendiculars from the vertices of a triangle upon
the opposite sides cut one another into segments that are
reciprocally proportional to each other.
53. The area of a triangle is equal to half the product of its
perimeter by the radius of the inscribed circle.
54. The perimeter of a triangle is to one side as the perpen
dicular from the opposite vertex is to the radius of the inscribed
circle.
55. The area of a square inscribed in a semicircle is equal to
two fifths of the area of the square inscribed in the circle.
56. The diagonals of any inscribed quadrilateral divide it
into two pairs of similar triangles.
57. To draw a line whose length is Vt^ in.
58. If two equivalent triangles are on the same base and the
same side of the base, any line cutting the triangles, and par
allel to the base, cuts off equal areas from the triangles.
59. To divide a given arc of a circle into two parts such that
their chords shall be in a given ratio.
60. The areas between two concentric circles may be found
by multiplying half the sum of the two circumferences by the
difference between the radii.
61. Find the length of the belt connecting two wheels of
the same size, if the radius of each wheel is 18 in., the distance
between the centers 6 ft., and 4 in. is allowed for sagging.
62. To construct a regular inscribed heptagon draftsmen
sometimes use for a side half the side of an inscribed equi
lateral triangle. Construct such a figure with the compasses,
and state whether the rule seems exact or only approximate.
APPENDIX
405. Subjects Treated. Of the many additionar subjects that
may occupy the attention of the student of plane geometry if
time permits, two are of special interest. These are Symmetry,
and Maxima and Minima.
406. Symmetric Points. Two points are said to be symmetric
with respect to a p)oi7it, called the center of symmetry, if this
third point bisects the straight line which joins the two points.
Two points are said to be symmetric with respect to an axis,
if a straight line, called the axis of symmetry, is the perpen
dicular bisector of the line joining them.
407. Symmetric Figure. A figure is said
to be symmetric with resjject to a pjoint, if
the point bisects every straight line drawn
through it and terminated by the boundary
of the figure.
A figure is said to be symmetric with
respect to an axis, if the axis bisects every /_
perpendicular through it and terminated
by the boundary of the figure.
Evidently this will be the case if one part coin
cides with another part when folded over the axis.
408. Two Symmetric Figures. Two figures
are said to be symmetric ivith respect to a
point or symmetric with respect to an axis,
if every point of each has a correspond
ing symmetric point in the other.
261
B'
262
APPENDIX TO PLANE GEOMETRY
Pkoposition I. Theorem
A quadrilateral that has tioo adjacent sides
and the other tivo sides equal, is symmetric
loith respect to the diagonal joining the vertices of the
angles formed hy the equal sides; and the diagonals
are perpendicular to each other.
409
equal
Given the quadrilateral ABCD, having AB equal to AD, and CB
equal to CD, and having the diagonals AC and BD.
To prove that the diagonal A C is an axis of symmetry^ and
that AC is A. to BD.
Proof. In the A yl/?C and ADC,
AB = AD, and CB = CD, Given
and AC = AC. Iden.
.'. A ABC is congruent to A ADC. § 80
.•.ZBAC = ZCAD, and ZACB=:ZDCA. §67
Hence, if A ^5C is turned on ^C as an axis until it falls
on A ADC, AB will fall on AD, CB on CD, and OB on OD.
.. the A ABC will coincide with the A ADC.
.'.AC will bisect every perpendicular drawn through it and
terminated by the boundary of the figure.
.*. ^C is an axis of symmetry. § 407
.. ^C is ± to BD, by § 406. Q.e.d.
SYMMETRY
263
Proposition II. Theorem
410. If a figure is symmetric with respect to tioo
axes j^erpendicular to each other, it is symmetric with
respect to their intersection as a center.

C
B
L
A
E
G
Y'
Given the figure ABCDEFGH, symmetric.with respect to the two
perpendicular axes XX^, YY', which intersect at O.
To prove that is the center of symmetry of the fiyure.
Proof. Let P be any point in the perimeter.
Draw PMQ J_ to YY\ and QNR _L to XX\
Then PQ is II to XX\ and QR is II to YY\
Draw PO, OR, and MN.
Then QN = NR.
{The figure is given as symmetric with respect to XX \)
But QN = MO.
.\NR = MO.
.'. RO is equal and parallel to NM.
In like manner, OP is equal and parallel to NM.
.'. ROP is a straight line.
.'.0 bisects PR, any straight line, and hence bisects every
straight line drawn through and terminated by the perimeter.
.*. O is the center of symmetry of the figure, by § 407. QE.d.
§227
§95
§407
§127
Ax. 8
§130
§94
264 APPENDIX TO PLANE GEOMETRY
EXERCISE 72
1. Draw a figure showing the number of axes of symmetry
possessed by a square.
2. Draw a figure showing the number of axes of symmetry
possessed by a regular hexagon.
3. Draw a figure showing six of the unlimited number of axes
of symmetry of a circle, and showing the center of symmetry.
4. Show by drawings that two congruent triangles may
be placed in a position of symmetry with respect to an axis.
In one of the drawings let a common side be the axis.
5. Show by a drawing that two congruent triangles may be
placed in a position of symmetry with respect to a center.
6. Two figures symmetric with respect to an axis are con
gruent.
7. Two figures symmetric with respect to a center are con
gruent.
8. Make a list of quadrilaterals that are symmetric with
respect to an axis.
9. Make a list of quadrilaterals that are symmetric with
respect to a center.
10. What kinds of regular polygons are symmetric with re
spect both to a center and to an axis ? Prove this for the hexagon.
11. A circle is symmetric with respect to its center as a
center of symmetry, and is also symmetric with respect to
any diameter as an axis.
12. An isosceles triangle is symmetric with respect to an axis,
and therefore the angles opposite the equal sides are equal.
13. Two tangents drawn to a circle from the same point are
symmetric with respect to an axis.
14. The four common tangents to two given circles form,
together with the circles, a figure symmetric with respect to
the line of centers as an axis.
MAXIMA AND MINIMA
265
411. Maxima and Minima. Among geometric magnitudes that
satisfy given conditions, the greatest is called the maximum, and
the smallest is called the minimum.
The plural of maximum is maxima, and the plural of minimum is
minima.
Among geometric magnitudes that satisfy given conditions, there may
be several equal magnitudes that are greater than any others. In this
case all are called maxima. .
Similarly there may be several minima magnitudes of a given kind.
412. Isoperimetric Polygons. Polygons which have equal
perimeters are called isoperimetric polygons.
If the circumference of a circle equals the perimeter of a polygon, the
circle and the polygon are said to be isoperimetric, and similarly for all
other closed figures in a plane.
Proposition III. Theorem
413. Of all triangles having two given sides, that in
which these sides include a right angle is the maximum.
P A F B
Given the triangles ABC and ABDy with AB and CA equal to
AB and DA respectively, and with angle BAC a right angle.
To prove that AABC> A ABD.
Proof. Prom D draw the altitude DP. § 227
Then DA > DP. § 86
But DA = CA. Given
.. CA > DP. Ax. 9
.'.A ABC >AABD, by § 327. Q.e.d.
266 APPENDIX TO PLANE GEOMETRY
Proposition IV. Theorem
414. Of all isoj^erimetric triangles having the same
base the isosceles triangle is the maximum.
B'
yi
./'
'/>
/
^;^
/h"
/ 1
/
HQ
Fig. 1
Fig. 2
§135
§215
§97
Given the triangles ABC and ABC having equal perimeters, and
having AC equal to BC, and AC not equal to BC\
To prove that A ABC > A ABC.
Proof. Produce AC to B', making CB' = AC.
Draw BB' and C'B', and draw C(2 II to AB.
Then since AC = CB', .\ BQ = QB'.
And since CA = CB = CB', ..ZB 'BA is a rt. Z.
.. CQ is _L to BB'.
C cannot lie on AB', for if it could, then CC'+C'B would
equal CB, which is impossible. Post. 1
Then since AC iCB' <AC' \ C'B', §112
.. AC\CB<AC'\C'B'. Ax. 9
.'. AC'\C'B<AC'{C'B'. Ax. 9
.. C'B < C'B'. Ax. 6
.. C ' cannot lie on CQ, for then C 'B would equal C 'B'. § 150
C'cannot lie above CQ (Fig. 1), for C"i>", which < C'P + P^',
would be less than C'B, which equals C'P + PB'.
.'. C must lie below CQ, as in Eig. 2.
.. A ABC > A ABC, by § 327. Q.e.d.
MAXIMA AND MINIMA 267
Proposition V. Theorem
415. Of all polygons with sides all given hut one, the
maximum can he inscribed in the semicircle ivhich has
the undetermined side for its diameter.
c
Given ABCDE^ the maximum of polygons with sides AB^ BC^
CDj DEj having the vertices A and E on the line MN.
To prove that ABODE can he inscribed in the semicircle
having EA for its diameter.
Proof. From any vertex, as C, draw CA and CE.
The A ACE must be the maximum of all A having the sides
CA and CE, and the third side on MN ; otherwise, by increas
ing or diminishing the Z EC A, keeping the lengths of the sides
CA and CE unchanged, but sliding the extremities A and E
along the line MN, we could increase the A A CE, while the
rest of the polygon would remain unchanged; and therefore
we could increase the polygon. But this is contrary to the
hypothesis that the polygon is the maximum polygon.
Hence the A ACE is the maximum of triangles that have
the sides CA and CE.
Therefore the Z. A CE is a right angle. § 413
Therefore C lies on the semicircle having EA for its
diameter. § 215
Hence every vertex lies on this semicircle.
That is, the maximum polygon can be inscribed in the semi
circle having the undetermined side for its diameter. q.e.d.
268
APPENDIX TO PLANE GEOMETRY
Proposition VI. Theorem
416. Of all polygons ivitJi given sides, one that can
he inscribed in a circle is the maximum.
Given the polygon ABCDE inscribed in a circle, and the polygon
A^B'C'D'E' which has its sides equal respectively to the sides of
ABCDE ^ but which cannot be inscribed in a circle.
To prove that ABCDE > A'B'C'B'U'.
Proof. Draw the diameter AP^ and draw CP and PD.
Upon CD' as a base, construct the A C'P'D' congruent to the
ACPD, and draw ^'P'.
Since, by hypothesis, a O cannot pass through all the vertices
of A'B'C'P'D'E', one or both of the parts A'P'D'E', A'B'C'P'
cannot be inscribed in a semicircle.
Neither A'P'D'E' or A'B'C'P' can be greater than its corre
sponding part. § 415
{Of all polygons with sides all given but one, the maximum can be inscribed
in the semicircle which Jias the undetermined side for its diameter.)
Therefore one of the parts A'P'D'E', A'B'C'P' must be less
than, and the other cannot be greater than, the corresponding
part of ABCPDE.
.. ABC PDE> A'B'C'P'D'E'.
Take from the two figures the congruent A CPD and C'P'B'.
Then ABCDE >A'B'C'D'E', by Ax. 6. Q.e.d.
MAXIMA AND MINIMA
Proposition VII. Theorem
269
417. Of isoperwietric polygons of a given niiinhe?' of
sides, the maximum is equilateral.
Given the polygon ABCDEF, the maximum of isoperimetric
polygons of n sides.
To prove that the polygon ABCDEF is equilateral.
Proof. Draw A C.
The A ABC must be the maximum of all the A which are
formed upon AC with a perimeter equal to that of A ABC.
Otherwise a greater A A PC could be substituted for AABC^
without changing the perimeter of the polygon.
But this is inconsistent with the fact that the polygon
ABCDEF is given as the maximum polygon.
.*. the A ABC is isosceles. § 414
.\AB = BC.
Similarly BC = CD, CD = DE, and so on.
.'. the polygon ABCDEF is equilateral. q.e.d.
418. Corollary. The maximum of isoperimetric polygons
of a given number of sides is a regular polygon.
For the maximum polygon is equilateral (§ 417), and can be inscribed
in a circle (§ 416). Therefore the maximum polygon is regular (§ 365).
270 APPENDIX TO PLANE GEOMETRY
Proposition VIII. Theorem
419. Of isoperwietric regular polygons, that ivhich
has the greatest nimiber of sides is the maximimi.
Y<
Given the regular polygon P of three sides, and the isoperimetric
regular polygon P' of four sides.
To prove that P'>P.
Proof. Draw CX from C to any point X in ^jB.
Invert the A AXC and place it in the position XCY, letting
X fall at C, C at X, and yl at F.
The polygon XBCY is an irregular polygon of four sides,
which by construction has the same perimeter as P' and the
same area as P.
Then the regular polygon P' of four sides is greater than
the isoperimetric irregular polygon XBCY of four sides. § 418
That is, a regular polygon of four sides is greater than the
isoperimetric regular polygon of three sides.
In like manner, it may be shown that P ' is less than the iso
perimetric regular polygon of five sides, and so on. q.e.d.
Discussion. We may illustrate this by the case of an equilateral tri
angle and a square, eacji with the perimeter.^. In the triangle the base
is I p, the altitude I p Vs, and the area J^p2 Vs, or about 0.048 p^. In the
square the base and altitude are each i p, and the area is ^^ p^, or 0.0625 jp^.
The area of the polygon is therefore increasing as we increase the number
of sides.
Since the limit approached by the perimeters is a circle, we may infer
that of all isoperimetric plane figures the circle has' the greatest area.
MAXIMA AND MINIMA
271
Proposition IX. Theorem
420. Of regular polygons having a given area, that
which has the greatest nuviber of sides has the minimum
2Jerimeter.
Given the regular polygons P and P' having the same area, P'
having the greater number of sides.
To prove that the perimeter of P > the perimeter of P'.
Proof. Construct the regular polygon P" having the same
perimeter as P \ and the same number of sides as P.
Denote a side of P by s, and a side of P " by s".
§419
Given
Ax. 9
§374
Then P'>P".
But P = P'.
.. P>P".
But P:P" = s^:s"\
.\s>s". Ax. 6
.'. the perimeter of P > the perimeter of P". Ax. 6
. But the perimeter of P ' = the perimeter of P ". Const.
.'. the perimeter of P > the perimeter of P', by Ax. 9. q.e. d.
Discussion. We may illustrate this, as on page 270, by the case of
an equilateral triangle and a square, each with area a^. The side of
the square is a, and the perimeter 4 a. The area of the equilateral tri
angle is I s2 Vs. Therefore ^ s^ Vs = a^^ or ^ s</d = a. Now </s = VVs ;
hence we have V3 = 1.734,and v V3= Vl. 73 = 1.3 + • Hence is x 1.3 = a,
and s = 1.5 a, and the perimeter of the triangle is 4.5 a. Therefore the
perimeter of the square is less than that of the triangle.
272 APPENDIX TO PLANE GEOMETRY
EXERCISE 73
Maxima and Minima
1. Of all equivalent parallelograms that have equal bases,
the rectangle has the minimum perimeter.
2. Of all equivalent rectangles, the square has the minimum
perimeter.
3. Of all triangles that have the same base and the same
altitude, the isosceles has the minimum perimeter.
4. Of all triangles that can be inscribed in a given circle, the
equilateral is the maximum and has the maximum perimeter.
5. To inscribe in a semicircle the maximum rectangle.
6. Find the area of the maximum triangle inscribed in a
semicircle whose radius is 3 in.
7. Of all polygons of a given number of sides that can
be inscribed in a given circle, that which is regular has the
maximum area and the maximum perimeter.
8. Of all polygons of a given number of sides that can be
circumscribed about a given circle, that which is regular has
the minimum area and the minimum perimeter.
9. In a given line required to find a point such that the
sum of its distances from two given points on the same
side of the line shall be the minimum. a
How does AP \ PB compare with A'B? and ^
this with A'X^ XB ? and this with AX+ XB ? J^
This is the problem of a ray of light from 4 to
the mirror CD, and reflected to B.
10. To divide a given line into two 1'
segments such that the sum of the squares on these segments
shall be the minimum.
11. To divide a given line into two segments such that their
product shall be the maximum.
EECREATIONS 273
421. Recreations of Geometry. The following simple puzzles
and recreations of geometry may serve the double purpose of
adding interest to the study of the subject and of leading the
student to exercise greater care in his demonstrations. They
have long been used for this purpose and are among the best
known puzzles of geometry.
EXERCISE 74
1. To prove that every triangle is isosceles.
Let ABC be a A that is not isosceles.
Take CP the bisector of ZACB, and ZP the ± bisector of AB.
These lines must meet, as at P, for otherwise c
they would be II, which would require CP to be _L j/f\
to AB, and this could only happen if A ABC were Y/ / \x
isosceles, which is not the case by hypothesis. >: ^'''''\ /
From P draw PX ± to BC and PY ± to C^, and ^ ^^^:'  "><^ ^
draw PA and PB. '' ^'
Then since ZP is the ± bisector of AB, .. PA = PB.
And since CP is the bisector of ZACB, .. PX = PY.
.. the rt. A PBX and PA Y are congruent, and BX = AY.
But the rt. A PXC and PYC are also congruent, and .. XC = YC.
Adding, we have BX + XC = AY \ YC, or BC = AC.
.. A ABC is isosceles even though constructed as not isosceles.
2. To prove that part of an angle equals the whole angle.
Take a square ABCD, and draw MM'P, the ± bisector of CD. Then
Jfilf'P is also the ± bisector of ylB. D M c
From B draw any line BX equal to AB.
Draw BX and bisect it by the _L NP.
Since BX intersects CD, Js to these lines can
not be parallel, and must meet as at P.
Draw PA, PB, PC, PX, and PB.
Since MP is the ± bisector of CD, PD=PC.
Similarly PA = PB, and PD = PX. X'fi^''''
.. PX = PD = PC. K
But BX — BC by construction, and PB is common to A PBX and PBC.
.: A PBX is congruent to A PBC, and Z XBP = Z CBP.
.: the whole Z XBP equals the part, Z CBP,
274
APPENDIX TO PLANE GEOMETRY
3. To prove that part of an angle equals the whole angle.
Take a right triangle ABC, and con
struct upon tlie hypotenuse BC an equi
lateral triangle BCD, as shown.
On CD lay off CP equal to CA.
Through X, the midpoint of AB,
draw PX to meet CB produced at Q.
Draw QA.
Draw the ± bisectors of QA and
QP, as YO and ZO. These must meet
at some point because they are ± to
two intersecting lines.
Draw OQ, OA, OP, and OC.
Since is on the ± bisector of QA, .. OQ = OA.
Similarly OQ = OP, and .. OA = OP.
But CA = CP, by construction, and CO = CO.
.: AA OC is congruent to A POC, and ZACO = Z PCO.
4. To prove that part of a line equals the whole line.
Take a triangle ABC, and draw CP ± to ^B.
From C draw CX, making Z^CX = ZB.
Then A ABC and ACX are similar.
.. AABC:AACX=BC^:CX^.
Furthermore A ABC: A ACX = AB: AX.
.'. BC" "
or BC^:AB=CX
But
A' P
CX^ = AB:AX,
^:AX.
BC^ = AC'^+ AB^
and
CX'
AC^ + AX^
f;2
AC'{AB"2ABAP
AC^+AX^2AX
2AB.AP,
2AXAP.
AP
AB
AX
^2
AC
AB
AC
^ AB2AP = —— + AX2AP.
AX
AC'
AB
AX
ACT
AX
AB,
AC'ABAX AC' ABAX
AB
AX
AB= AX.
RECREATIONS
275
5. To show geometrically that 1=0.
Take a square that is 8 units on a side, and cut it into three parts,
4, 7?, C, as shown in the righthand figure. Fit
these parts together as in the lefthand figure.
Now tlie square is 8 units on a side, and therefore
contains 8 x 8, or 64, small squares, while the rec
tangle is 13 units long and 5 units high, and there
fore contains
5 X 13, or 65,
small squares.
But the two
figures are each made up of J. + J5+ C
(Ax.ll), and therefore are equal (Ax.8).
64 we have 1 z= (Ax. 2).
+ _4__,__(_.
65 = 64, and by subtracting
6. To prove that any point on a line bisects it.
Take any point P on AB. q
On AB construct an isosceles A ABC, having /*\
AC=BC; and draw PC. / \\
Then in A APC and PBC, we have / \ \
ZA=ZB, § 74 / \ \
AC=BC, Const. / \ \
and PC = PC. Iden. "^ J' ^
Three independent parts (that is, not merely the three angles) of one
triangle are respectively equal to three parts of the other, and the tri
angles are congruent ; therefore AP = BP (§ 67).
7. To prove that it is possible to let fall two perpendiculars
to a line from an external point.
Take two intersecting © with centers and (7.
Let one point of intersection be P, and draw the diameters PA and PD.
Draw AD cutting the circumferences at B
and C. Then draw PB and PC.
Since Z PC A is inscribed in a semicircle,
it is a right angle. In the same way, since
ZDBP is inscribed in a semicircle, it also is
a right angle.
.. PB and PC are both X to AD.
276
APPENDIX TO PLANE GEOMETRY
8. To prove that if two opposite sides of a quadrilateral are
equal the figure is an isosceles tra]oezoid.
Given the quadrilateral ABCD, with BC = DA.
To prove that ^J5 is I to DC.
Draw MO and NO, the ± bisectors of AB and
CD, to meet at 0.
li AB and DC are parallel, the proposition is already proved.
U AB and DC are not parallel, then MO and NO will meet at 0, either
inside or outside the figure. Let be supposed to be inside the figure.
Draw OA, OB, OC, OD.
Then since OM is the ± bisector of AB, .. OA = OB.
Similarly
and
Also,
and
Similarly
and
OD = OC.
But DA is given equal to BC.
.'. AAOD is congruent to A BOC,
ZDOA = ZBOC.
rt. ii OCN and ODN are congruent,
ZNOD = ZCON.
rt. A AMO and BMO are congruent,
ZAOM = ZMOB.
.: Z NOD + Z DO A \ZAOM=Z CON + Z BOC + Z MOB,
or Z NOM = Z MON = a st. Z.
Therefore the line MON is a straight line, and hence AB is II to DC.
D ^^^,
If the point is outside the quadrilateral, as
In the second figure, the proof is substantially the
same.
For it can be easily shown that A/
ZDONZDOAZAOM 1/
= Z NOC  Z BOC  Z MOB, ^
which is possible only if
ZDON=ZDOM,
or if ON lies along OM.
But that the proposition is not true is evident from the
third figure, in which BC = DA, but ^Z^ is not II to DC.
O
'N \
M
HISTORY OF GEOMETRY 277
422. History of Geometry. The geometry of very ancient
peoples was largely the mensuration of simple areas and
volumes such as is taught to children in elementary arithmetic
today. They learned how to find the area of a rectangle,
and in the oldest mathematical records that we have there is
some discussion of triangles and of the volumes of solids.
The earliest documents that we have, relating to geometry,
come to us from Babylon and Egypt. Those from Babylon
were written about 2000 b.c. on small clay tablets, some of
them about the size of the hand, these tablets afterwards
having been baked in the sun. They show that the Baby
lonians of that period knew something of land measures, and
perhaps had advanced far enough to compute the area of a
trapezoid. For the mensuration of the circle they later used,
as did the early Hebrews, the value tt = 3.
The first definite knowledge that we have of Egyptian math
ematics comes to us from a manuscript copied on papyrus, a
kind of paper used about the Mediterranean in early times.
This copy was made by one Aahmesu (The Moonborn), com
monly called Ahmes, who probably flourished about 1700 b.c.
The original from which he copied, written about 2300 b.c,
has been lost, but the papyrus of Ahmes, written nearly four
thousand years ago, is still preserved and is now in the British
Museum. In this manuscript, which is devoted chiefly to frac
tions and to a crude algebra, is found some work on mensu
ration. Among the curious rules are the incorrect ones that
the area of an isosceles triangle equals half the product of
the base and one of the equal sides; and that the area of a
trapezoid having bases &, h\ and nonparallel sides each equal
to (X, is ^a(b\b'). One noteworthy advance appears however.
Ahmes gives a rule for finding the area of a circle, substan
tially as follows : Multiply the square on the radius by (V)^
which is equivalent to taking for ir the value 3.1605. Long
before the time of Ahmes, however, Egypt had a good working
278 APPENDIX TO PLANE GEOMETRY
knowledge of practical geometry, as witness the building of
the pyramids, the laying out of temples, and the digging of
irrigation canals.
From Egypt and possibly from Babylon geometry passed to
the shores of Asia Minor and Greece. The scientific study of
the subject begins with Thales, one of the Seven Wise Men
of the Grecian civilization. Born at Miletus about 640 b.c, he
died at Athens in 548 b.c. . He spent his early manhood as a
merchant, accumulating the wealth that enabled him to spend
his later years in study. He visited Egypt and is said to have
learned such elements of geometry as were known there. He
founded a school of mathematics and philosophy at Miletus,
known as the Ionic School. How elementary the knowledge
of geometry then was, may be understood from the fact that
tradition attributes only about four propositions to Thales,
substantially those given in §§ 60, 72, 74, and 215 of this book.
The greatest pupil of Thales, and one of the most remark
able men of antiquity, was Pythagoras. Born probably on the
island of Samos, just off the coast of Asia Minor, about the
year 580 b.c, Pythagoras set forth as a young man to travel.
He went to Miletus and studied under Thales, probably spent
several years in study in Egypt, very likely went to Babylon,
and possibly went even to India, since tradition asserts this
and the nature of his work in mathematics confirms it. In
later life he went to southern Italy, and there, at Crotona, in
the southeastern part of the peninsula, he founded a school
and established a secret society to propagate his doctrines.
In geometry he is said to have been the first to demonstrate
the proposition that the square on the hypotenuse of a right
triangle is equivalent to the sum of the squares on the other
two sides (§ 337). The proposition was known before his time,
at any rate for special cases, but he seems to have been the
first to prove it. To him or to his school seems also to have
been due the construction of the regular pentagon (§§ 397, 398)
HISTORY OF GEOMETRY 279
and of the five regular polyhedrons. The construction of the
regular pentagon requires the dividing of a line in extreme
and mean ratio (§ 311), and this problem is commonly assigned
to the Pythagoreans, although it played an important part in
Plato's school. Pythagoras is also said to have known that six
equilateral triangles, three regular hexagons, or four squares,
can be placed about a point so as just to fill the 360°, but that
no other regular polygons can be so placed. To his school is
also due the proof that the sum of the angles of a triangle
equals two right angles (§ 107), and the construction of at
least one starpolygon, the starpentagon, which became the
badge of his fraternity.
For two centuries after Pythagoras geometry passed through
a period of discovery of propositions. The state of the science
may be seen from the fact that (Enopides of Chios, who
flourished about 465 b.c, showed how to let fall a perpendicu
lar to a line (§ 227), and how to construct an angle equal to a
given angle (§ 232). A few years later, about 440 b.c, Hippoc
rates of Chios wrote the first Greek textbook on mathematics.
He knew that the areas of circles are proportional to the squares
on their radii, but was ignorant of the fact that equal central
angles or equal inscribed angles intercept equal arcs.
About 430 B.C. Antiphon and Bryson, two Greek teachers,
worked on the mensuration of the circle. The former attempted
to find the area by doubling the number of sides of a regular
inscribed polygon, and the latter by doing the same for both in
scribed and circumscribed polygons. They thus substantially
exhausted the area between the circle and the polygon, and
hence this method was known as the Method of Exhaustions.
During this period the great philosophic school of Plato
(429348 B.C.) flourished at Athens, and to this school is due
the first systematic attempt to create exact definitions, axioms,
and postulates, and to distinguish between elementary and
higher geometry. At this time elementary geometry became
280 APPENDIX TO PLANE GEOMETRY
limited to the use of the compasses and the unmarked straight
edge, which took from this domain the possibility of con
structing a square equivalent to a given circle (" squaring the
circle"), of trisecting any given angle, and of constructing
a cube with twice the volume of a given cube (" duplicating
the cube"), these being the three most famous problems of
antiquity. Plato and his school were interested in the socalled
Pythagorean numbers, numbers that represent the three sides
of a right triangle. Pythagoras had already given a rule to
the effect that i(m''{lf = 7n^ + i(m^iy. The school of
Plato found that [(^my\iy = m^i[(^mylf. By giving
various values to vi, different numbers will be found such that
the sum of the squares of two of them is equal to the square of
the third.
The first great textbook on geometry, and the most famous
one that has ever appeared, was written by Euclid, who taught
mathematics in the great university at Alexandria, Egypt,
about 300 B.C. Alexandria was then practically a Greek city,
having been named in honor of Alexander the Great, and
being ruled by the Greeks.
Euclid's work is known as the "Elements," and, as was the case
with all ancient works, the leading divisions were called books,
as is seen in the Bible and in such Latin writers as Caesar
and Vergil. This is why we speak of the various books of
geometry today. In this work Euclid placed all the leading
propositions of plane geometry as then known, and arranged
them in a logical order. Most subsequent geometries of any im
portance since his time have been based upon Euclid, improving
the sequence, symbols, and wording as occasion demanded.
Euclid did not give much solid geometry because not much
was known then. It was to Archimedes (287212 b.c), a
famous mathematician of Syracuse, on the island of Sicily,
that some of the most important propositions of solid geometry
are due, particularly those relating to the sphere and cylinder.
HISTORY OF GEOMETRY 281
He also showed how to find the approximate value of tt by a
method similar to the one we teach today (§ 404), proving
that the real value lies between 3} and 3}^. Tradition says
that the sphere and cylinder were engraved upon his tomb.
The Greeks contributed little more to elementary geometry,
although Apollonius of Perga, who taught at Alexandria be
tween 250 and 200 b.c, wrote extensively on conic sections; and
Heron of Alexandria, about the beginning of the Christian era,
showed that the area of a triangle whose sides are a, b, c, equals
V6'(.s — a) (s — b) (s — c), where s = ^(a\b^c) (see p. 211).
The East did little for geometry, although contributing
considerably to algebra. The first great Hindu writer was
Aryabhatta, who was born in 476 a.d. He gave the very
close approximation for tt, expressed in modern notation as
3.1416. The Arabs, about the time of the Arabian Nights Tales
(800 A.D.), did much for mathematics, translating the Greek
authors into their own language and also bringing learning
from India. Indeed, it is to them that modern Europe owes
its first knowledge of Euclid. They contributed nothing of
importance to geometry, however.
Euclid was translated from the Arabic into Latin in the
twelfth century, Greek manuscripts not being then at hand, or
being neglected because of ignorance of the language. The
leading translators were Athelhard of Bath (1120), an English
monk who had learned Arabic in Spain or in Egypt ; Gerhard
of Cremona, an Italian monk ; and Johannes Campanus, chap
lain to Pope Urban IV.
In the Middle Ages in Europe nothing worthy of note was
added to the geometry of the Greeks. The first edition of
Euclid was printed in Latin in 1482, the first one in English
appearing in 1570. Our symbols are modern,  and — first
appearing in a German work in 1489; = in Recorde's "Whet
stone of Witte" in 1557; > and < in the works of Harriot
(15601621); and x in a publication by Oughtred (15741660).
282 APPENDIX TO PLANE GEOMETRY
423. Notation used in Formulas. Following the general cus
tom, small letters represent numerical values, large letters rep
resent points. The following abbreviations have been used in
this book as consistently as the circumstances would allow, the
context telling which abbreviation is intended :
a = area, apothem I = length.
a, b, c = sides oiAABC. m = median,
a' = projection of a. 2^ — perimeter.
b, h' = bases. ^ r = radius.
c = circumference. s = semiperimeter of A,
d = diameter, diagonal. 1 (a + Z» f c).
h = height, altitude. ir = 3.1416, or about 3}.
424. Formulas for Line Values. The following are the most
important formulas in line values :
Right triangle, a' ib'' = c" (§ 337).
Any triangle, a^ ^ b'' ± 2 aV = c" {^% 341, 342).
Circle, c = 2 irr = ird (§ 385).
Radius of circle, v — c^2'Tr.
Equilateral triangle, ^^ = \b y 3.
Diagonal of square, d = b V2 (§ 339).
Side of square, b = Va.
425. Areas of Plane Figures. The following are the formulas
for the areas of the most important j^lane figures :
Rectangle, bh (§ 320).
Square, b'' (§ 320).
Parallelogram, bh (§ 322).
Triangle, ^^bh(^S25),^s(sa)(sb){sc).
Equilateral triangle, ^ b'^ v 3.
Trapezoid, V^(^ + ^') (§^29).
Regular polygon, ^ ap (§ 386).
Circle, ^rc = 7rr2(§§388, 389).
INDEX
Page
Acute angle 16
Acute triangfe 26
Adjacent angles 7
Alternation, proportion by .152
Altitude 59
Analytic proof . . 80, 140, 141
Angle 6
acute 16
at center of regular polygon 227
central 93
complement of .... 18
conjugate of 18
exterior 51
inscribed 115
measure of 18
oblique 16
obtuse 16
plane 6
reentrant . . . ^ . . 08
reflex 16
right 7, 16
sides of 6
size of 6, 17
straight 16
supplement of 18
vertex of 6
Angles, adjacent 7
alternateinterior . . . 47
complementary .... 18
conjugate 18
corresponding 26
equal 6
Page
Angles, exterior . . , . 47, 51
exteriorinterior .... 47
generation of 17
interior 47, 51
made by a transversal . . 47
of a polygon 68
of a triangle 7
supplementary . . . . 18
vertical 18
Antecedents 151
Apothem 227
Arc 7,93
major 93
minor 93
Area of circle 115
of irregular polygon . . 199
of surface 191
Attack, methods of . . 140, 145
Axiom 21
Axioms, list of 22
Axis of symmetry .... 261
Base . 7, 32, 59
Bisector 6, 74
Broken line 5
Center of circle 7
of regular polygon . . . 227
of symmetry 261
Central angle 93
Chord 95
Circle 7,93
283
284
INDEX
Page
Circle, arc of . . .
. . 7, 93
area of ... .
. . .115
as a limit . . .
. 114, 237
as a locus . . .
... 93
center of . . .
... 7
central angle of .
... 93
chord of . . .
... 95
circumference of .
... 7
circumscribed . .
. . .114
diameter of . .
. . 7, 93
inscribed . . .
... 114
radius of . . .
. . 7, 93
secant to . . ,
. 102, 177
sector of . . .
. . . 115
segment of . . .
. . .115
tangent to . . .
. . .102
Circles, concentric .
. . .104
escribed ....
... 137
tangent ....
. . .107
Circumcenter . . .
. . 78, 136
Circumference . .
... 7
Circumscribed circle
. . .114
Circumscribed polygor
1 . . .114
Commensurable magn
tudes . 112
Common measure
. . .112
Common tangents
. . .109
Complement . . .
... 18
Composition, proportic
m by .153
Concave polygon . .
... 68
Concentric circles .
... 104
Concurrent lines . .
.... 77
Congruent ....
. . 26,68
Conjugate ....
... 18
Consequents . . .
. . .151
Constant ....
. . .114
Continued proportion
. . .151
Continuity, principle c
)f . . 125
Converse propositions
. . 35, 95
Converse theorems, la
w of .95
Convex polygon . .
. , . 68
Page
Corollary 21
Corresponding angles ... 26
Corresponding lines .... 165
Corresponding sides . . 26, 165
Curve 5
Curvilinear figure .... 6
Decagon 68
Degree .... .^ ... 18
Determinate cases .... 140
Diagonal 59, 68
Diameter 7, 93
Difference of magnitudes . . 17
Dimensions 2
Discussion of a problem 126, 140
Distance 42
Division, harmonic .... 161
proportion by 154
Dodecagon ....... 68
Drawing figures . . . 8, 29, 84
Equal angles 6
Equal lines 5
Equiangular polygon ... 68
Equiangular triangle ... 26
Equilateral polygon .... 68
Equilateral triangle .... 26
Equivalent figures . . . . 191
Escribed circles 137
Excenter 137
Exterior angles . . . . 47, 51
Extreme and mean ratio . . 184
Extremes ....... 151
Figure 4
curvilinear 5
geometric ...... 4
plane 4
rectilinear 6
symmetric 261
INDEX
285
Figures, equivalent .
isoperimetric . .
symmetric . . .
Foot of perpendicular
Formulas ....
Fourth proportional .
Generation of angles
of magnitudes . .
Geometric figure . .
Geometry . . . .'
History of . . .
Harmonic division .
Heptagon . . . .
Hexagon . . . .
History of Geometry
Homologous angles .
Homologous lines
Homologous sides
Hypotenuse . . .
Hypothesis . . . .
Impossible cases
Incenter . . .
Incommensurable magnitudes
Incommensurable ratio
Indeterminate cases .
Indirect proof .
Inscribed angle . .
Inscribed circle .
Inscribed polygon
Instruments
Interior angles . .
Inversion, proportion
Isoperimetric polygons
Isosceles trapezoid .
Isosceles triangle . .
by
Limit
Page page
. 191 Limits, principle of ... . 115
. 265 Line 3,5
. 261 broken 5
7 curve 5
. 282 of centers 107
. 151 segments of .... 5, 161
straight 5
. 17 Lines, concurrent . . . .77
4,17 corresponding. , .• . , 165
4 equal 5
4 oblique 16
. 277 parallel 46
perpendicular 7
product of 194
transversal of 47
Loci, solutions by .... 143
Locus 73
proof of 74
Magnitudes 3
bisectors of 6
commensurable .... 112
constant 114
differences of 17
generation of . . . . 4, 17
incommensurable . . .112
sums of 17
variable 114
Maximum 265
Mean proportional .... 151
Means 151
Measure 112
. 8 angle 18
47, 51 common 112
. 153 numerical .... 112, 117
. 265 Median 77
. 59 Methods of attack . . 140, 145
. 26 of proof. . 35,77,80,8.3,84
Minimum 265
114, 237 Multiple 112
161
277
26
165
26
42
30
140
137
112
113
140
83
115
114
114
286
INDEX
Page
Nature of proof 25
of solution 126
Negative quantities .... 125
Nonagon 68
Numerical measure . . 112, 117
Oblique angle 16
Oblique lines 16
Obtuse angle 16
Obtuse triangle 26
Octagon . . 68
Optical illusions 15
Parallel lines 46
Parallelogram 59
Pentagon 68
Pentadecagon 246
Perigon 18
Perimeter 7, 68
Perpendicular 7
Perpendicular bisector ... 74
Pi(7r) 238
value of 249
Plane 3
angle 6
geometry 4
Point 3
of contact .... 102, 107
Polygon 68
angles of 68
apothem of regular . . .227
area of 191, 199
center of regular .... 227
circumcenter of ... . 136
circumscribed 114
concave 68
convex 68
diagonal of .... 59, 68
equiangular 68
equilateral 68
Page
Polygon, incenter of . . . 137
inscribed 114
perimeter of 68
radius of regular .... 227
regular . 68, 227
sides of 68
vertices of 68
Polygons, classified . . 68, 114
congruent 68
isoperimetric 265
mutually equiangular . . 68
mutually equilateral . . 68
similar 165
Positive quantities .... 125
Postulate . 21
of parallels 46
Postulates, list of .... 23
Principle of continuity . . . 125
of limits 115
Problem 21, 126
how to attack . . . 140, 145
Product of lines 194
Projection 205
Proof, methods of 35, 77, 80, 83, 84
nature of 25
necessity for 15
Proportion 151
continued 151
nature of quantities in a .155
Proportional, fourth . . . .151
mean 151
reciprocally 177
third 151
Proposition 21
Pythagorean theorem . . . 204
Quadrilateral 59, 68
Quadrilaterals classified . . 59
Radius 7, 93
of regular polygon . . . 227
INDEX
287
Page
Ratio 112
extreme and mean . . .184
incommensurable . . .113
of similitude 165
Recreations of Geometry . . 273
Rectangle 59
Rectilinear figure 5
Reductio ad absurdum ... 83
Reentrant angle 68
Reflex angle 16
Regular polygon . . . .68, 227
Rhomboid 59
Rhombus 59
Right angle 7, 16
Right triangle 26
Scalene triangle 26
Secant 102, 177
Sector 115
Segment of a circle . . . .115
of a line 5, 161
Semicircle 93, 115
Sides, corresponding ... 26
of angle 6
of triangle 7
of polygon . . ... . 68
Similar parts of circles . . . 239
Similar polygons 165
Similitude, ratio of ... . 165
Size of angle 6, 17
Solid 2
Solid geometry 4
Solution, nature of ... . 126
Square 26
Straight angle 16
Straight line 5
Subtend 93, 95
Sum of magnitudes . . . . 17
Superposition 35
Supplement 18
Page
Surface 3, 191
Symmetric figures .... 261
Symmetry 261
Synthetic proof . . 35, 77, 140
Tangent .... 102, 107, 109
Tangent circles 107
Terms of a proportion . . .151
Theorem . 21
Third proportional . . . .151
Transversal 47
Trapezium 59
Trapezoid 59
Triangle 7, 68
acute 26
altitude of 59
angles of 7
base of 7, 59
circumcenter of . . . 78, 136
equiangular 26
equilateral 26
excenter of 137
incenter of 78, 137
isosceles 26
obtuse 26
right 26
scalene 26
sides of 7
vertices of 7
Triangles classified .... 26
Unit of measure 112
of surface 191
Variable 114
Vertex of angle 6
of isosceles triangle . . 59
Vertical angles 18
Vertices of a polygon ... 68
of a triangle 7
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